UNIVERSITY  OF  CALIFORNIA 
AT  LOS  ANGELES 


r  of  CALIFORNIA 

AT 

ANGELES 
UKRARY 


METHODS 


IN 


Written  Arithmetic, 

BY 

7*'*  7 

JOHN  W.  COOK,  A.  M., 

PROFESSOR  OF  MATHEMATICS  IN 

ILLINOIS  STATE  NORMAL  UNIVERSITY. 


REVISED  EDITION. 

CHICAGO: 
A.  Fi.AN'AG\N,  PUBLISHER. 


COPYRIGHTED,  BY  A.  FLANAGAN,  1883. 


QA 


CONTENTS. 


PAGE. 

Preface,       .......        5 

Definitions,  Notation,  Numeration,     ...  9 

Addition,     ....  ...       19 

Subtraction,       .  .....  25 

Multiplication,        '.  .  ....       29 

Division,  ......  37 

Factoring,    .  .  .  .  .  .  .46 

Cancellation,      .,-....  56 

Multiples,    .  .  .  .  .  .  64 

Fractions,  ......  68 

Addition  of  Fractions,       .  ...  .  .74 

Division  and  Partition  of  Fractions,     ...  80 

The  Method  of  Finding  the  Part  which  one  Number  is  of 
Another,  .  .  .  .  .  .87 

Decimal  Fractions,  •  «  •  •  93 


Contents. 

I  AGE. 

Compound  Numbers,           .                                       ,  100 

Measures  of  Length,     .             .  109 
Longitude  of  Time,              .              .             .                            .118 

Percentage,        ....  I26 

Loss  and  Gain,        ,  .  . 

Commission,       .  . 

Stock  Investments,              .....  145 

Interest,              ...»  I5° 

Partial  Payments,                 .             .                                         •  158 

Present  Worth,               .             .            .            •             •  1°° 

Bank  Discount,        ......  163 

Exchange,          ...... 

Equation  of  Payments,       .            .             .             •            •  179 
Evolution                                      .            •             •             .182 


PREFACE. 


It  has  long  been  the  conviction  of  the  writer  of  the 
following  pages  that  the  subject  of  arithmetic,  as  it  is 
too  often  studied,  yields  a  smaller  return  for  the  time 
devoted  to  it,  than  any  other  branch  included  in  the 
curriculum  of  our  common  schools.  Pupils  begin  the 
study  of  numbers  at  the  threshold  of  their  school  life 
and  continue  it  for  from  six  to  ten  years ;  yet  a  large 
proportion  of  them  have  but  little  appreciation  of  the 
principles  of  the  science,  and  their  work  is,  conse- 
quently, almost  entirely  mechanical. 

A  subject  that  receives  so  large  a  share  of  the 
pupils' time  should  accomplish  something  more  in  the 
direction  of  real  education  than  to  give  the  power  to 
make  simple  calculations  with  tolerable  accuracj 

If  the  subject  is  left  without  having  given  the 
child  the  ability  to  define  with  sharp  discrimination. 


Preface. 

to  analyze  with  fluency,  and  to  give  the  reasons  for 
the  operations  involved,  the  work  must  be  regarded 
as,  at  best,  but  a  partial  success. 

This  little  volume  doesniotpi^tend  to  be  an  ex- 
haustive treatise.  Its  purpose  is  simply  to  present 
detailed  methods  of  work  in  the  more  familiar  parts 
of  the  subject,  to  suggest  forms  of  analysis  that  seem 
to  be  reasonably  acceptable,  to  state  as  clearly  as 
possible  the  leading  principles  involved,  and  to  urge 
the  necessity  of  thought  work  in  the  arithmetic  class. 
No  attempt  is  made  to  furnish  drill  problems.  It  is 
expected  that  the  book  will  be  used  in  connection 
with  some  regular  text-book  as  a  supplementary  work 
in  analysis,  etc.  It  owes  its  appearance  to  the  re- 
quests of  many  normal  students  who  desired  to  see 
certain  features  of  their  work  in  arithmetic  put  into 
permanent  and  accessible  form. 

With  the  hope  that  it  may  be  helpful  to  some,  at 
least,  of  my  fellow  teachers,  it  is  herewith  submitted. 

NORMAL,  ILLINOIS. 


METHODS 

IN 

WRITTEN   ARITHMETIC. 


CHAPTER  I. 


ARITHMETIC  is  the  science  which  treats  of  number, 
its  nature  and  properties,  the  methods  of  writing, 
reading,  combining  and  resolving  it,  and  of  its  appli- 
cation to  practical  affairs. — Olney. 

Number  is  that  which  answers  the  question,  "How 
many  ?  " 

Number  is  one,  or  is  formed  by  the  repetition  of 
ones. 

Ones  are  either  Absolute  or  Relative. 

The  Absolute  unit  is  without  limitation. 


TO  Methods  in  Written  Arithmetic. 

Relative  units  are  units  formed  by  uniting  Abso- 
lute units,  or  by  the  separation  of  an  Absolute  unit 
into  equal  parts. 

Numbers  are  either  Integral  or  Fractional. 

An  Integral  uumber  is  a  number  that  is  composed 
of  one  or  more  Integral  units. 

An  Integral  unit  is  an  Absolute  unit  or  a  Relative 
unit  formed  by  uniting  Absolute  units. 

A  Fractional  number  is  a  number  that  is  com- 
posed of  one  or  more  Fractional  units. 

A  Fractional  unit  is  a  Relative  unit  that  is  formed 
by  the  separation  of  an  Absolute  unit  into  equal 
parts. 

But  three  operations  are  possible  with  numbers. 
They  may  be  expressed,  united,  and  separated. 

Numbers  may  be  expressed  by  characters,  or 
orally. 

Notation  treats  of  the  methods  of  expressing  num- 
bers by  means  of  characters. 

Numeration  treats  of  the  methods  of  reading  num 
bers  that  are  expressed  by  figures. 

There  are  three  methods  of  Notation:  (i.)  The 
Word  Method;  (2.)  The  Letter  Method;  (3.)  The 
Figure  Method. 


Methods  in  Written  Arithmetic.  II 

The  Word  Method  is  used  in  general  correspond- 
ence, essays,  and  similar  productions. 

The  Letter  Method  is  usually  employed  in  paging 
prefaces  and  appendixes;  in  numbering  volumes, 
lessons,  and  chapters;  in  the  titles  of  monarchs;  on 
the  dials  of  time-pieces,  etc. 

It  employs  the  printed  letters  I,  V,  X,  L,  C,  D,  M, 
and  the  dash. 

I  =  i,  V=5,  X=io,  L=5o,  C=ioo,  D  =  5oo, 
M=iooo.  Both  the  capitals  and  lower-case  letters 
are  used.  The  scale  is  five  and  two;  thus  five 
I's=one  V;  two  V's=one  X,  etc. 

LAWS. 

1.  Repeating  a  letter  repeats  its  value. 

2.  When  a  letter  is  placed  before  one  of  greater 
value,  the  two  express  a  number  equal  to  the  differ- 
ence of  their  values. 

3.  When  a  letter  is  placed  after  one  of  greater 
value,  the  two  express  a  number  equal  to  the  sum  of 
their  values. 

4.  When  a  letter  is  placed  between  two,  each  of 
greater  value,  the  three  express  a  number  equal  to 
the  difference  of  the  values  of  the  less  and  of  the  sura 
of  the  two  greater. 


12  Methods  in  Written  Arithmetic. 

5.  Placing  a  dash  over  a  letter  multiplies  its  value 
by  one  thousand. 

Pupils  of  the  grade  implied  by  this  work  should 
learn  to  read  and  write  readily  at  least  as  far  as 
M  M,  and  usually  much  farther. 

The  Figure,  or  Arabic  Method  of  notation  em- 
ploys ten  characters.  They  are :  T,  2,  3,  4,  5,  6,  7, 
3,  9,  o. 

These  characters  are  called  digits,  from  the  Latin 
digit/is,  meaning  finger. 

The  first  nine  are  called  significant  figures.  They 
have  two  values :  an  absolute  or  shape  value,  and  a 
relative  or  place  value. 

The  zero  is  used  to  fill  vacant  places  between 
some  significant  figure  and  the  decimal  point. 

The  decimal  point  is  used  to  fix  the  order  of  abso- 
lute units.  In  doing  so  it  also  determines  the  names 
of  relative  units. 

A  scale  is  the  statement  of  the  number  of  units  in 
each  order  required  to  make  one  of  the  next  higher. 

A  decimal  system  of  numbers  is  one  in  which  the 
scale  is  ten. 

There  are  two  systems  of  Numeration:  The 
French  and  the  English. 


Methods  in  Written  Arithmetic.  13 

The  former  is  the  one  in  common  use  in  this 
country.  It  groups  figures  into  periods  of  three 
orders  each,  beginning  at  the  right.  It  enables  the 
reader  to  tell  how  many  are  expressed  by  the  suc- 
cession of  figures. 

The  right-hand  place,  or  order,  in  each  group  is 
called  units'  order;  the  second,  tens';  the  third, 
hundreds'. 

Each  group  is  read  as  if  standing  alone ;  then  the 
name  of  the  period  is  spoken,  and,  at  the  close,  the 
kind  of  units  numbered. 

Notation  and  Numeration  are  taught  together. 

Pupils  should  be  able  to  read  and  write  with  per- 
fect readiness  all  numbers  from  quadrillions  to  bil- 
lionths.  When  readiness  has  been  acquired  in  this 
work,  pupils  can  extend  it  at  liberty. 

Teachers  have  noticed  the  trouble  that  pupils 
experience,  in  writing,  in  determining  whether  their 
work  is  correct.  They  stand  with  crayon  in  one 
hand  and  eraser  in  the  other,  and  alternately  write, 
"numerate"  and  erase.  The  writing  should  be  done 
rapidly,  no  erasing  should  be  permitted,  and  the 
instant  the  writing  is  completed  the  pupil  should  turn 
from  the  board.  In  order  to  do  this,  certain  knowl- 
edge is  necessary. 


14  Methods  in  Written  Arithmetic. 

1.  The  pupil  must  know  the  order  of  periods, 
from  left  to  right. 

2.  He  should  know  the  number  of  each  period 
at  the  left  of  the  point,  and  of  each  order  at  the 
right. 

The  number  of  the  period  can  be  fixed  easily  by 
observing  the  meaning  of  the  prefixes  in  the  names. 

The  force  of  the  bi  in  billions  is  two;  of  the  tri  in 
trillions  is  three,  and  so  on. 

Observe  that: 

The  number  of  any  period  above  millions  is  two 
more  than  the  meaning  of  the  prefix  in  the  name  of 
the  period. 

The  numbers  of  the  orders  at  the  right  of  the  point 
should  be  thoroughly  memorized,  so  that  the  pupil 
can  speak  the  number  if  the  name  is  given,  or  vice 
versa. 

3.  The  instant  that  a  period  is  filled  it  should  be 
followed  by  a  comma,  unless  it  is  units'  period. 

Suppose  a  pupil  is  asked  to  write  17  trillions,  286 
millions,  521.  He  writes  17  and  follows  it  with  a 
comma.  He  remembers  that  trillions'  period  is  the 
fifth,  and  millions'  is  the  third,  so  he  writes  three 
zeroes  and  a  comma,  then  286  and  a  comma.  Re- 
membering that  thousands'  period  is  the  second,  he 


Methods  in  Written  Arithmetic.  15 

writes  three  zeroes  and  a  comma,  and  then  521  and 
the  period.  In  the  early  work  in  writing  it  is  wise 
to  insist  upon  the  invariable  use  of  the  period. 

If  the  pupil  is  asked  to  write  624  millionths,  he 
should  remember  that  the  4  must  stand  in  the  sixth 
order  at  the  right;  hence  he  writes  the  period,  three 
zeroes  and  624.  He  has  no  need  to  "numerate," 
and  it  should  not  be  permitted. 

In  reading  numbers  the  word  and  should  not  be 
used  except  with  expressions  that  lie  on  both  sides 
of  the  point,  and  then  only  when  read  as  mixed 
numbers.  Thus,  56.693  should  be  read  as  fifty-six 
and  six  hundred  ninety-three  thousandths  (a 
mixed  number),  or  as  fifty-six  thousand  six  hundred 
ninety-three  thousandths  (a  simple  number). 

EXERCISES. 

i.  To  ascertain  whether  the  pupils  have  memo- 
rized the  numbers  of  the  periods  and  orders,  Dicta- 
tion Exercises,  like  the  following,  may  be  given: 
With  the  class  at  the  board,  or  supplied  with  slates, 
the  teacher  rapidly  names  the  periods,  thus :  millions, 
sextillions,  thousands,  trillions,  etc.,  and  'he  pupils 
write,  3,  8,  2,  5,  etc. 


16  Methods  in  Written  Arithmetic. 

2.  Give  a  similar  exercise  for  orders  at  the  right 
of  the  point. 

3.  In  writing  large  numbers,  let  the  teacher  read 
a  single  period  and  wait  for  the  pupil  to  write,  then 
another,  etc. 

4.  Vary  exercise    (3)   by  requiring  the   class  to 
face  the  teacher  while  he  speaks  the  number  once, 
then  at  a  signal  class  turns  and  writes,  putting  in 
commas,  doing  no  erasing  nor  "numerating,"  and 
"facing"  immediately  when  the  work  is  completed. 

5.  Instead  of  exercise  (2)  give  the  following  oc- 
casionally :     Give  the  number  to  be  written  without 
naming  the  kind  and  without  permitting  the  voice  to 
fall.    When  it  is  written  speak  the  denomination  and 
the  pupils  put  the  period  into  its  proper  position. 

6.  Have  the  pupils  "  write  without  the  crayon," 
thus:     They  read  "63  trillions,  589  millions,  274," 
and  then  say:  63,  comma,  three  zeroes,  comma,  589 
comma,  three  zeroes,  comma,  274,  period."    Use  the 
same  exercise  in  writing  decimals. 

7.  In  preparing  numbers  for  reading,  the  names 
of  the  periods  should  be  spoken  as  the  commas  are 
placed,  and  the  reading  should  be  begun  as  soon  as 
the  "pointing"  is  completed. 


Methods  in  Written  Arithmetic.  17 

It  is  of  the  utmost  importance  that  pupils  acquire 
the  habit  of  picturing  the  exact  appearance  of  an  ex- 
pression before  making  any  efforts  at  writing.  Thus, 
in  writing  decimal  fractions  they  should  think,  i.  Of 
the  number  of  figures  necessary  to  express  the  how 
many  there  are  in  the  number;  2.  Where  the  right- 
hand  figure  must  stand  in  order  that  the  denomina- 
tion may  be  expressed  properly;  3.  How  many 
zeroes  must  be  placed  between  the  decimal  point 
and  the  first  significant  figure  to  make  the  right-hand 
figure  stand  in  the  proper  order. 

If,  for  example,  they  are  told  to  write  5608  ten- 
millionths,  they  should  notice,  (i)  that  four  figures 
are  necessary  to  express  5608;  (2)  that  the  8  must 
stand  in  the  seventh  order  at  the  right  of  the  point 
to  make  the  number  ten-millionths,  and  (3)  that  three 
zeroes  must  stand  between  the  point  and  the 
5 ;  hence  they  write,  decimal  point,  three  zeroes, 
5608. 

CAUTIONS. 

1.  Names  of  orders  are  often  given  incorrectly. 
Say  "ten-thousandths,"  not  "tens  of  thousandths;" 
"hundred-million ths,"  not  hundreds  of  millionths." 

2.  When  the  word  period  or  order  is  understood 


1 8  Methods  in  Written  Arithmetic. 

after  the  noun,  the  noun  should  be  written  with  the 
apostrophe:  billionths' period;  thousandths' order. 

3.  See  that  all  board  work  is  scrupulously  neat. 

4.  Allow  no  one  to  use  an  eraser  without  permis- 
sion. 

The  English  System  of  Numeration  has  passed  out 
of  use  in  this  country,  but  the  names  are  retained  in 
the  French  System,  although  their  meaning  has  been 
lost. 

The  English  System  employs  six  orders  in  a 
period.  The  name  of  the  first  period  is  units;  the 
second,  millions;  the  third,  billions,  etc. 

Million  is  derived  from  mille,  a  thousand.  It  is  a 
thousand  thousands. 

Billion  is  contracted  from  bis,  twice,  and  million. 
It  is  the  square  of  a  million. 

Trillion  is  from  tres,  three,  and  million.  It  is  the 
third  power  of  a  million;  and  so  following. 

In  the  French  System  these  terms,  except  million, 
are  used  arbitrarily,  a  billion  being  a  thousand  mill- 
ions, and  a  trillion  a  thousand  billions. 


CHAPTER    II. 


NUMBERS  are  united  in  two  ways:  (i.)  By  Addi- 
tion; (2.)  By  Multiplication.  The  second  may  be 
regarded  as  a  special  case  in  Addition. 

The  grade  of  the  work  discussed  herein  should  not 
be  forgotten.  The  pupils  are  supposed  fo  have  done 
an  amount  of  work  represented  by  the  First  Book  in 
Number.  It  should  be  remembered  that  proficiency 
of  the  pupils  will  depend  upon  the  amount  of  work 
that  each  does  for  himself,  hence  it  should  be  the 
aim  of  the  teacher  to  keep  every  member  of  the  class 
as  busy  as  possible  during  the  recitation. 

Try  a  teachers'  Institute  on  a  problem  in  Addition 
which  involves  a  dozen  three-place  numbers,  and  ten 
per  cent,  of  the  results  will  be  incorrect.  The  rea- 
son is  obvious — they  have  not  the  technical  knowl- 
edge necessary  for  rapid  and  correct  work.  The 
memory  must  be  stored  with  facts,  and  in  this  case 
each  of  the  facts  is  the  sum  of  two  numbers. 


2o  Methods  in  Written  Arithmetic. 

The  ground  facts  to  be  thoroughly  mastered  are 
the  sums  of  any  two  numbers  each  of  which  is  less 
than  ten.  These  are  contained  in  the  Addition 
tables  of  any  First  Book.  When  a  pupil  knows  that 
the  sum  of  8  and  9  is  17,  he  can  readily  learn  that 
the  sum  of  9  and  any  two-place  number  ending  in  8 
has  7  for  its  right-hand  digit,  and  that  the  number  of 
tens  is  increased  by  one. 

Addition  is  the  process  of  finding  the  sum  of  two 
or  more  like  numbers. 

The  sum  of  two  or  more  numbers  is  a  number 
which  contains  as  many  units  as  all  of  the  numbers 
taken  together. 

EXERCISES. 

Each  pupil  should  be  provided  with  a  slate  and 
pencil,  or  their  equivalent. 

i.  The  teacher  stands  before  the  class  and  pro- 
nounces eight  or  ten  one-place  numbers  as  rapidly  as 
the  class  can  work  "mentally."  When  his  voice  falls, 
the  pupils  write  the  result.  The  same  exercise  is 
repeated  until  six  or  eight  problems  have  been  per- 
formed, the  teacher  meanwhile  keeping  a  list  of  the 
several  sums.  The  results  should  then  be  examined. 
This  exercise  need  not  occupy  more  than  three  min« 


Methods  in  Written  Arithmetic,  21 

utes.     Pupils  should  not  be  permitted  to  see  each 
other's  work. 

2.  Send  the  class  to  the  board.    Give  several  dic- 
tation exercises  in  two-place,  or  larger  numbers,  the 
pupils  writing  all  the  numbers. 

In  order  that  they  may  get  no  aid  from  each  other, 
let  them  number  in  order  and  assign  one  problem  to 
the  odd  numbers  and  another  to  the  even. 

Permit  no  erasing  and  no  changing.  Hold  all 
responsible  for  the  first  result.  Insist  that  the  work 
shall  be  scrupulously  neat.  .Carry  the  work  to  the 
right  of  the  point  as  in  the  writing  suggested  in  the 
previous  chapter. 

3.  Write  upon  the  board  a  line  Of  figures  and  one 
figure  below  the  line,  thus : 

3896745693 
8 

Require  one  pupil  to  give  the  sum  of  8  and  3; 
another  of  8  and  8,  and  so  on  around  the  class. 
This  should  go  very  rapidly. 

The  figures  may  be  written  as  follows,  and  another 
be  placed  within,  thus : 


23  Methods  in  Written  Arithmetic. 

3 

4  8 

6  9 

9  * 

797 

3  4 

«  3 

5  8 
i 

By  this  means  the  exercise  can  go  on  indefinitely 
without  interruption.  Pupils  should  not  be  permit- 
ted to  name  the  numbers  added,  but  should  give  the 
results  only. 

4.  Introduce  competitive  exercises.  These  must 
be  used  with  care,. as.  some  pupils  are  easily  confused 
or  discouraged.  Multiply  exercises  until  the  pupils 
are  stimulated  to  do  considerable  practice-work  out 
of  the  class. 

LANGUAGE   WORK. 

Results  in  arithmetic  are  of  no  account  unless  cor- 
rect. Accuracy  is  the  first  essential.  Equal  defi- 
niteness  should  be  secured  in  the  language  work.  He 
misses  the  best  culture  that  the  study  of  arithmetic 
should  afford  who  leaves  it  without  the  power  to  talk 
clearly  and  pointedly. 


Methods  in  Written  Arithmetic.  2$ 

Definitions  should  be  memorized  with  extreme 
c«re.  Analyses  should  be  given  promptly  and  accu- 
rately, and  without  the  aid  of  suggestive  questions. 
The  pupils  should  be  required  to  go  through  a  com- 
plete explanation  while  the  teacher  simply  listens. 
Much  patient  labor  is  required  to  reach  such  a  result, 
but  no  teacher  should  be  satisfied  with  less.  Pupils 
maybe  aided  at  first  by  judicious  questions,  but  help 
should  be  withdrawn  as  soon  as  possible,  and  they 
should  be  required  to  work  alone. 

Give  them  a  chance  to  do  something  for  themselves. 

Faulty  expressions,  too  many  of  which  are  found 
in  some  text-books,  should  be  corrected.  Figures 
are  not  numbers  any  more  than  a  photograph  is  a 
man,  hence  they  cannot  be  added.  Pupils  talk 
about  writing  figures  "under  each  other."  How 
many  writings  of  each  would  be  required  to  place  2 
and  5  under  each  other?  "  The  sum  of  9  and  7 
would  be  1 6,"  is  often  heard.  What  is  implied  in 
such  an  expression? 

A  statement  somewhat  like  the  following  should 
be  secured: 

For  convenience  I  write  the  numbers  with  units 
of  the  same  order  in  the  same  column.  Beginning 
with  the  lowest  order  I  find  the  sum  of  the  numbers 


24  Methods  in  Written  Arithmetic. 

expressed  in  each  column.  If  the  sum  in  any  case 
is  nine  or  less,  I  express  it  under  the  column  added; 
if  more  than  nine,  I  reduce  it  to  the  next  higher 
order,  writing  the  remainder,  if  any,  under  the  col- 
umn added,  and  adding  the  reduced  number  to  the 
first  number  expressed  in  the  next  column,  etc. 

QUESTIONS. 

Is  it  necessary  to  write  the  numbers  as  above? 
Why  more  convenient?  Is  it  necessary  to  begin 
with  the  lowest  order?  When  is  i«~  equally  conve- 
nient to  begin  with  the  highest?  When  is  it  more 
convenient  to  begin  with  the  lowest?  Why  is  it 
more  convenient? 

The  so-called  "proofs"  are  only  checks  against 
errors.  Addition  in  the  opposite  direction  is  the 
simplest  test  of  accuracy. 

An  ordinary  problem  in  addition  includes  a  num- 
ber of  problems  in  each  of  the  fundamental  pro- 
cesses. See  that  the  pupils  recognize  this  fact, 

Repetition  is  the  secret  of  success.  He  adds  well 
who  has  added  a  great  deal. 


CHAPTER  III. 


THERE  are  two  methods  of  resolving  numbers: 
I.  Subtraction,  and  2.  Division. 

Subtraction  is  the  process  of  separating  a  number 
into  two  parts,  one  of  which  is  given  for  the  purpose 
of  finding  the  other  part. 

The  definition  may  be  verified  by  performing  a 
problem  with  objects. 

"If  John  had  eight  sticks  and  gave  three  of  them 
to  Thomas  how  many  had  he  left?" 

The  problem  frequently  assumes  a  different  form  : 

"If  John  had  eight  sticks  and  Thomas  three,  John 
had  how  many  more  than  Thomas?" 

The  solution  obviously  is  the  same  as  in  the  form- 
er problem.  Separate  John's  sticks  into  two  parts, 
one  of  which  shall  contain  as  many  sticks  as  Thomas 
had.  The  remainder  is  found  in  the  same  manner  as 
before, 


z6  Methods  in  Written  Arithmetic. 

Where  pupils  have  acquired  a  knowledge  of  sub« 
traction  by  performing  the  operations  with  objects, 
there  will  be  no  confusion  of  ideas. 

DEFINITIONS   OF  TERMS. 

The  Minuend  is  a  number  that  is  to  be  separated 
into  two  parts,  one  of  which  is  given  for  the  purpose 
of  finding  the  other. 

The  Subtrahend  is  the  given  part  of  the  minuend. 

The  Remainder  is  the  required  part  of  the  min- 
uend. 

The  operation  consists  in  the  separation  of  the 
minuend  into  subtrahend  and  remainder.  If  they 
be  re-united  of  course  the  minuend  will  result  from 
the  operation. 

The  simplest  form  of  a  problem  in  subtraction  is 
that  in  which  each  term  of  the  minuend  equals  or 
exceeds  the  corresponding  term  of  the  subtrahend. 
When  the  problem  is  presented  in  this  form  it  is,  of 
course,  immaterial  where  the  subtraction  begins;  e. 
g:  8256 — 4132=?  Problems  rarely  present  this 
form  however,  yet  all  must  assume  it  before  the  sub- 
traction can  be  performed.  There  are  two  methods 
of  changing  the  ordinary  problems  into  the  simplest 
form.  The  simpler  is  called  the  "changed  minuend" 


Methods  in  Written  Arithmetic.  27 

method,  and  consists  in  so  changing  the  form  of  the 
minuend,  without  altering  its  value,  that  each  of  its 
terms  shall  equal  or  exceed  the  corresponding  term 
of  the  subtrahend.  In  ordinary  work  the  change  is 
made  as  the  necessity  arises,  but  with  beginners  the 
wiser  plan  is  to  make  all  the  changes  at  the  same 
time.  Take  the  following  as  an  illustration  ;  7253 — 
2879=  ?  Teach  the  pupils  that  7253  equals  6  thous- 
ands, n  hundreds,  14  tens  and  13  units..  The  prob- 
lem then  assumes  the  following  form : 

6     n     14     13 

2879 

When  this  is  understood  let  the  changes  in  the 
minuend  be  made  as  the  necessity  arises. 

The  explanation  will  assume  a  form  sustantially 
like  the  following : 

"Since  I  cannot  separate  3  units  into  two  parts, 
one  of  which  is  9  units,  I  reduce  i  ten  to  units:  i 
ten  =  10  units;  10  units  -(-  3  units  =  13  units.  If 
13  units  be  separated  into  two  parts,  one  of  which 
is  9  units,  the  remainder  will  be  4  units,  etc." 

CAUTIONS. 

i.  Do  not  permit  the  pupils  to  "borrow."  "Bow- 
rowing"  with  no  purpose  of  returning  is  a  bad  habit. 


28  Methods  in  Written  Arithmetic. 

2.  Where   analyses  like   the   above   are   written, 
allow  no  false  statements  of  equality.     Forms  like 
the  following  are  common:    T    ten  =  10    units   + 
3  units  =13   units  —  9    units  =  4    units.       From 
the  foregoing   we    must    conclude    that  i   ten  =  4 
units.     Require  the    writing    to    be    truthful,    even 
at  the  expense  of  time  and    material,     i  ten  =  10 
units.     10   units+3  units=i3  units.      13    units — 9 
units=4  units.    To  prevent  false  forms  in  writing, 
insist  that  the  oral  analysis  shall  be  equally  explicit. 

3.  To  furnish   additional   tests,  assign  a  written 
analysis  for  home  work.     Insist  that  the  papers  shall 
be  uniform  in  size  and  folding,  that  they  shall  be 
written  with  ink  and  that  they  shall  be  neat. 

A  second  method  of  changing  a  problem  into  the 
form  for  subtraction  is  called  "The  Old  Method." 
It  rests  upon  the  axiom  that  if  two  numbers  be 
equally  increased  their  difference  remains  the  same. 

Illustration:     7253  —  2879  =  ? 

Since  I  cannot  separate  3  units  into  two  parts,  one 
of  which  is  9  units,  I  add  10  units  to  the  units'  terra 
of  the  minuend.  3  units  +  I0  units  =  13  units. 
13  units  —  9  units  =  4  units.  Since  I  have  added 
10  units  to  the  minuend  I  must  add  10  units  to  the 
subtrahend.  (Why?)  10  units  =  i  ten.  7  tens  -j- 
i  ten  =  8  tens,  etc. 


CHAPTER  IV. 


As  has  been  said,  Multiplication  is  a  method  of 
aniting  numbers.  It  may  be  regarded  as  a  special 
case  in  addition,  in  which  the  numbers  to  be  united 
are  equal.  3x4=?  simply  means,  what  is  the 
sum  of  three  fours?  The  facts  of  the  multiplication 
table  are  fixed  in  the  memory,  and  the  operation 
consists  in  appealing  to  these  memories. 

DEFINITIONS. 

1.  Multiplication   is   a   short   method   of  uniting 
equal  numbers. 

2.  The  Multiplicand  is  one  of  two  or  more  equal 
numbers  that  are  to  be  united. 

3.  The  Multiplier  is  the  number  of  equal  numbers 
that  are  to  be  united. 

4.  The  Product  is  the  sum  of  two  or  more  equal 
numbers  that  have  been  united. 


30  Methods  in  Written  Arithmetic. 

An  understanding  of  the  literal  meaning  of  the 
above  terms  will  be  helpful. 

From  the  definition  of  number,  it  follows  that  the 
distinction  ordinarily  made  between  Abstract  and 
Concrete  numbers  has  no  foundation  in  fact. 

Although  number  is  purely  abstract,  a  knowledge 
of  it  can  be  acquired  only  by  the  use  of  objects; 
hence  the  terms  may  be  found  convenient;  with  this 
in  mind,  we  may  enunciate  the  following 

PRINCIPLES. 

1.  The   Multiplicand  may  be  either   abstract   or 
concrete. 

2.  The  Multiplier  is  always  abstract. 

3.  The  Product  is  like  the  Multiplicand. 

That  many  pupils  who  are  studying  written  arith- 
metic have  no  clear  idea  of  the  nature  of  multiplica- 
tion is  demonstrated  by  the  fact  that  when  asked  to 
show  with  objects,  the  meaning  of  3  X  5,  they  will 
take  three  objects  in  one  hand  and  five  in  the  other. 
3X5  does  not  mean  three  fives  to  them.  They  have 
studied  figures,  not  numbers.  The  use  of  "times"  is, 
perhaps,  responsible,  in  part,  for  such  a  mental  con- 
dition. The  word  is  never  a  necessity.  It  is  some- 


Methods  in  Written  Arithmetic.  31 

times  a  convenience,  but  in  many  cases,  probably, 
serves  to  hide  knowledge.  It  is  wiser  to  omit  it  un- 
til the  nature  of  multiplication  is  understood. 

LANGUAGE   WORK. 

Do  not  under-estimate  the  value  of  the  language 
drill.  Insist  upon  clear,  smooth  and  grammatical 
explanations.  A  form  similar  to  the  following  may 
profitably  be  insisted  upon : 

(i.)     489X364=? 

489  X  364=  9X  364+80  X  364+400  X  364. 

9X4  units=36  units=3  tens  and  6  units. 

9X6  tens=54  tens.     54  tens+3  tens=57  tens= 

5  hundreds  and  7  tens,  etc. 

80X364=8X10X364.     10X364=364  tens. 

8X4  tens  =  32  tens,  etc.     400  X  364  =  4X  100  X 

364.     100  X  364=364  hundreds. 

4X4  hundreds=i6  hundreds,  etc. 

The  above  form  will  be  found  simpler  than  the 
more  common  form. 

(2.)  Pupils  often  blunder  in  explaining  the  multipli- 
cation by  the  factors  of  a  number.  This  should  not 
be  permitted.  24X36=6x4x36,  4X36=144. 
6x  144=864. 

(3,)  Pupils  should  acquire  facility  in  analyzing 
problems  involving  concrete  numbers. 


32  Methods  in  Written  Arithmetic. 

What  is  the  cost  of  80  acres  of  land  at  $24  an 
acre? — Analysis:  If  each  acre  cost  $24,  80  acres 
cost  8oX$24,  or  $1920. 

Why  not  24  X  80?     Why  is  each  better  than  one? 

(4.)  A  very  useful  exercise,  and  one  that  prepares 
pupils  for  future  work,  and  emphasizes  the  principles 
of  multiplication,  consists  in  changing  the  order  of 
factors  and  explaining  carefully. 

For  example,  make  24  the  multiplier  in  the  above 
problem. 

It  is  evident  from  the  third  principle  that  there 
can  be  no  product  which  shall  be  dollars  until  we 
have  a  multiplicand  that  is  dollars.  Analysis:  If 
each  acre  cost  but  one  dollar,  80  acres  would  cost 
80  dollars.  Since  each  acre  cost  24  dollars,  80  acres 
cost  24x80  dollars,  or  1920  dollars. 

The  ability  to  make  a  supposition  and  show  what 
the  result  would  be,  and  then  so  modify  the  result 
as  to  make  it  agree  with  given  conditions,  is  very 
valuable,  not  only  in  arithmetic,  but  in  all  subsequent 
mathematical  study. 

(5.)  "To  multiply  by  10,  100,  1000,  etc.,  annex 
as  many  zeroes  to  the  multiplicand  as  there  are  in 
the  multiplier." 


Methods  in  Written  Arithmetic.  33 

The  above  statement  is  a  very  common  one.  Is 
it  correct?  If  I  annex  one  zero  to  the  right  of  78. 
I  have  78.0.  Have  I  multiplied  78  by  10? 

It  may  be  said  that  usually  the  decimal  point  is 
not  placed  at  the  right  of  an  integer.  But  suppose 
the  number  had  been  7.8,  we  should  then  have  had 
7.80.  Is  it  not  a  better  statement  to  say:  To  multi- 
ply by  10  make  the  number  stand  one  order  farther 
to  the  left,  or  (if  the  decimal  point  is  used),  remove 
the  decimal  point  one  order  to  the  right? 

(6.)  There  are  several  short  methods  of  multipli- 
cation, and  with  some  of  these,  at  least,  pupils  should 
become  familiar.  They  serve  a  double  purpose. 
They  are  of  considerable  practical  importance,  and 
the  analysis  of  them  prepares  the  way  for  explana- 
tions of  processes  in  fractions. 

In  this  list  should  be  included  multiplication  by 
aliquot  parts  of  one  hundred.  The  pupils  should 
Itarn  the  following  "aliquots"  : 

One-half;  the  thirds;  the  fourths;  one-sixth  and 
five-sixths;  one,  three,  five  and  seven-eighths.  The 
twelfths  and  sixteenths  may  be  added  where  time 
permits. 


34  Methods  in  Written  Arithmetic. 

ILLUSTRATION. 

Multiply  864  by  37^. 

864X37% 


32400 

Insist  upon  a  particular  form^  ^rcnnj*trt«Si,  and 
neatness. 

Explanation.     37^  =  ^tofvoo. 

Were  the  multiplier  100,  tne  product  would  be 
86400,  obtained  by  moving  rhe  multiplicand  two  or- 
ders to  the  left.  Were  th*  multiplier  one-eighth  of 
100,  the  product  would  be  one-eighth  of  86400,  or 
10800.  Since  the  multiplier  is  three-eighths  of  100, 
the  product  is  three  times  10800,  or  32400. 

Require  the  pupils  to  write  analyses  similar  to  the 
above,  giving  especial  attention  to  paragraphing, 
capitals,  spelling  and  punctuation. 

(7.)  An  extension  of  this  work  is  advisable  where 
circumstances  are  favorable,  thus:  37^=375.  375 
of  any  kind  is  three-eighths  of  one  of  the  first  order 
ar  th<?  left  of  the  3.  3.75=^  of  10;  .375  =  ^  of  i; 


Methods  in  Written  Arithmetic.  35 

375  —  ^6  of  1000,  etc.    In  the  same  way,  read  8.3/4, 
.833/3;  1.875:  31250. 

Example.     Multiply  482  by  6875. 

Form.  482  x  6875 

4820000 


30*250 


Explanation.     6875  =|£  of  10000. 

Were  the  multiplier  10000,  the  product  would  be 
4,820,000,  obtained  by  making  the  number  stand  4 
orders  farther  to  the  left.  Were  the  multiplier  1-16 
of  10000,  the  product  would  be  1-16  of  4,820,000,  or 
301,250.  Since  the  multiplier  is  11-16  of  icooo,  the 
product  is  11X301,250,  or  3,313,750. 

The  methods  above  noted  form  a  large  part  of 
the  "stock-in-trade"  of  many  "lightning  calculators." 

(8.)  Many  text-books  give  a  short  method  for 
multipliers  which  are  near  a  power  of  ten. 

Multiply  87964  by  997. 

Form.  87964X997 

87964000 


263892 


87,700,108 
Explanation.     997  is  3  less  than  1000. 


36  Methods  in  Written  Arithmetic. 

Were  the  multiplier  1000,  the  product  would  be 
87,964,000,  obtained  by  making  the  number  stand 
three  orders  farther  to  the  left.  Since  the  multiplier 
is  3  less  than  1000,  the  product  is  3x87,964  less 
than  87,964,000. 

To  be  able  to  use  the  sixteenths  of  one  hundred 
readily  as  multipliers,  pupils  must  learn  the  first  nine 
multiples  of  sixteen. 

Give  drill  exercises  upon  them  until  they  are  mas- 
tered. Since  in  this  work  sixteen  is  to  be  used  as  a 
divisor,  the  following  exercise  will  be  found  advan- 
tageous. Write  a  large  number  on  the  board,  as  for 
example, 

258938756983- 

Let  the  pupil  name  (i)  the  largest  multiple  of  six- 
teen in  the  partial  dividend,  then  (2)  the  quotient, 
next  (3)  the  remainder,  and  finally  (4)  the  new  par- 
tial dividend. 

In  the  above  example  the  numbers  spoken  would 
be  (i)  sixteen;  (2)  one;  (3)  nine;  (4)  ninety-eight. 
The  process  would  then  be  repeated  until  the  close. 
When  facility  is  required  only  the  successive  terms 
of  the  quotient  should  be  spoken,  the  other  results 
being  thought  only. 


CHAPTER  V. 


Division  is  by  far  the  most  difficult  of  the  "funda- 
mental" processes.  This  is  due  to  the  fact  that  two 
processes  have  received  the  same  name.  1 8-^-3  is 
usually  read,  "Divide  18  by  3,"  yet  it  has  been  trans- 
lated "Separate  18  into  threes,"  and  "Find  one-third 
of  18,  that  is,  separate  18  into  three  equal  parts." 

The  processes  with  objects  are  far  from  identical. 
A  bundle  of  sticks  is  separated  into  threes  by  count- 
ing off  a  bundle  of  three,  and  continuing  the  opera- 
tion until  the  bundle  is  separated  into  smaller  bun- 
dles of  three  each.  The  second  operation  consists 
in  starting  three  smaller  bundles  with  one  in  each, 
and  increasing  them  equally  until  the  operation  is 
completed.  In  the  first  case  there  are  found  to  be 
6  bundles  of  three  each;  in  the  second,  each  of  the 
three  is  found  to  contain  6  sticks. 

It  seems  wiser  to  have  a  name  for  each  operation. 
If  the  term  Division  be  reserved  for  the  former,  the 


38  Methods  in  Writtey  Arithmetic. 

word  Partition  may  be  appropriately  used  for  the 
latter.  Perform  the  operation  with  objects  and  test 
the  accuracy  of  the  following 

DEFINITIONS. 

Division  is  the  process  of  separating  a  number 
into  equal  numbers  of  a  given  size  for  the  purpose  of 
finding  the  number  of  such  numbers  in  the  given 
number. 

The  Dividend  in  Division  is  the  number  to  be 
separated  into  equal  numbers  of  a  given  size. 

The  Divisor  in  Division  is  the  number  showing 
the  size  of  the  equal  numbers  into  which  the  Divi- 
dend is  to  be  separated. 

The  Quotient  in  Division,  is  the  number  which 
shows  the  number  of  equal  numbers  into  which  the 
Dividend  has  been  separated. 

The  Divisor  is  like  the  Dividend.  The  Quotient 
is  abstract. 

In  "proving"  Division,  which  becomes  the  multi- 
plier? Why? 

Partition  is  the  process  of  separating  a  number 
into  a  given  number  of  equal  numbers  for  the  pur- 
pose of  ascertaining  the  size  of  each. 


Methods  in  Written  Arithmetic.  39 

The  Dividend  in  Partition  is  the  number  which  is 
to  be  separated  into  a  given  number  of  equal  num- 
bers. 

The  Divisor  in  Partition  is  the  number  which 
shows  the  number  of  equal  numbers  into  which  the 
Dividend  is  to  be  separated. 

The  Quotient  in  Partition  is  the  number  which 
shows  the  size  of  the  equal  numbers  into  which  the 
Dividend  has  been  separated. 

In  "proving"  a  problem  in  Partition  which  be- 
comes the  multiplier?  Why? 

An  example  in  Division : 

If  15  apples  be  divided  into  piles  of  three  apples 
each,  how  many  piles  will  there  be? 

Analysis:     If  15   apples  be   divided   into  piles  of 

I 

three  apples  each  there  will  be  as  many  piles  as 
there  are  threes  of  apples  in  fifteen  apples.  There 
are  five  threes  of  apples  in  fifteen  apples :  hence 
there  will  be  five  piles. 

An  example  in  Partition : 

If  15  apples  be  divided  into  5  equal  piles,  how 
many  apples  will  there  be  in  each  pile? 

Analysis:  If  15  apples  be  divided  into  5  equal 
piles,  there  will  be  one-fifth  of  15  apples,  or  3  apples, 
in  each  pile. 


4O  Methods  in  Written  Arithmetic. 

The  following  is  a  briefer  statement  of  the  sub- 
ject : 

Division  and  Partition  are  processes  of  finding  a 
second  factor  when  the  product  and  one  factor  are 
given.  The  Divisor  is  the  given  factor.  The  Divi- 
dend is  the  given  product.  The  Quotient  is  the  re- 
quired factor.  When  the  given  factor  and  the  pro- 
duct are  alike  the  problem  falls  under  Division; 
when  unlike,  under  Partition.  It  is  wiser  to  confine 
this  statement  to  problems  in  which  the  quotient  is 
integral.  When,  in  performing  a  problem,  the  sev- 
eral products  and  remainders  are  written,  the  division 
is  said  to  be  Long;  when  these  are  not  written  the 
division  is  called  Short. 

- 

METHODS  OF  EXPLAINING  ABSTRACT  PROBLEMS. 

Division.  8634-7-9  (Read,  divide  8634  into  nines.) 
There  are  9  nines  in  86,  Avith  a  remainder  of  5. 
Since  the  86  is  hundreds  the  quotient  and  remainder 
are  hundreds.  5  hundreds=5o  tens.  50  tens-|-3 
tens  are  53  tens.  There  are  5  nines  in  53,  with  a 
remainder  of  8.  Since  the  53  is  tens,  the  quotient 
and  remainder  are  tens.  8  tens=8o  units.  80  units 
+  4  units=84  units.  There  are  9  nines  in  84,  with 


Methods  in   Written  Arithmetic.  4! 

a  remainder  of  3;  hence  in  8634  there  are  959  nines 
with  a  remainder  of  3. 

Partition.     8634-4-9  (Read,  find  1  of  8634.) 

1  of  86  hundreds=9  hundreds,  with  a  remainder 
of  5  hundreds.  5  hundreds=5o  tens.  50  tens+3 
tens=53  tens.  \  of  53  tens=5  tens,  with  a  remain- 
der of  8  tens.  8  tens=8o  units.  80  units  +  4  units 
=  84  units.  \  of  84  units =9  units,  with  a  remain- 
der of  9  units. 

Require  each  concrete  problem  to  be  read  prompt- 
ly and  accurately,  assigned  to  its  proper  case  and 
analyzed.  By  this  time  the  pupils  should  have  ac- 
quired the  ability  to  talk  with  decided  fluency.  Hes- 
itation and  stumbling  should  not  be  tolerated. 

Changing  problems  from  one  case  to  the  other 
will  be  found  to  be  a  profitable  exercise 

ILLUSTRATION. 

A  bought  a  farm  at  $25  an  acre,  paying  $2875  f°r 
it;  how  many  acres  did  it  contain? 

This  is  a  problem  in  Division,  because  Divisor  and 
Dividend  are  like  numbers. 

It  may  be  changed  to  Partition. 


42  Methods  in  Written  Arithmetic. 

Had  he  paid  but  one  dollar  an  acre,  $2875  would 
have  bought  2875  acres.  Since  he  paid  $25  an  acre, 
$2875  bought  only  ^  of  2875  acres,  or  115  acres.' 

A  bought  115  acres  of  land  for  $2875,  what  was 
the  price  an  acre? 

This  is  a  problem  in  Partition.     Why? 
It  may  be  changed  to  Division. 

Had  he  paid  only  one  dollar  an  acre,  115  acres 
would  have  cost  $115.  Since  the  115  acres  cost 
$2875,  ne  Paid  as  rnany  dollars  an  acre  as  there  aie 
115*5  of  dollars  in  $2875.  There  are  25  115*5  of 
dollars  in  $2875  ;  hence  he  paid  $25  an  acre. 

Division  by  aliquot  parts  of  100.  1000,  etc.,  furn- 
ishes a  good  opportunity  for  language  drill,  fits  the 
pupils  for  other  analyses,  and  supplies  a  short  method 
of  obtaining  results  in  many  problems 

ILLUSTRATION. 

Problem.—  86485^-87^  =  ? 


Form.  86485-=- 

864.85 

6918.80 


Methods  in  Written  Arithmetic.  43 


Explanation.     87^  =  ^3  of  100. 

Were  -the  divisor  100,  the  quotient  would  be 
864.85,  obtained  by  removing  the  number  two  or- 
ders to  the  right.  Were  the  divisor  ^  of  100,  the 
quotient  would  be  8x864.85,  or  6918.8.  Since  the 
divisor  is  fa  of  100,  the  quotient  is  \  of  6918.8,  or 
988.4. 

Do  not  overlook  what  may  be  called  the  "elocu- 
tion" of  an  oral  analysis. 

Pronunciation  should  be  distinct  and  accurate,  and 
contrasted  terms  should  be  properly  emphasized. 
Pupils  should  stand  erect  and  should  face  the  teacher 
or  the  class  instead  of  the  blackboard. 

It  is  convenient,  often,  to  simplify  a  problem  by 
dividing  by  the  factors  of  the  divisor. 

The  explanation  should  be  given  with  care.  Y1T  of 
576  =  ^  of  *%  of  576.  %  of  576  =  96.  X°f96  =  24- 

The  process  of  finding  the  true  remainder  is  often 
puzzling  to  pupils. 

Example.     8661-5-42  =  ? 

Into  how  many  groups  of  42  units  each  can  I  sep 
arate  8661? 

8661  ones=433o  twos,  with  a  remainder  of  one 
one.  4330  twos=i443  sixes,  with  a  remainder  of 


44 


Methods  in  Written  Arithmetic. 


one  two.  1443  sixes=2o6  forty-twos,  with  a  re- 
mainder of  one  six;  hence,  8661  ones  equals  204 
forty-twos,  with  a  remainder  of  one  six,  one  two  and 
one  one,  or  nine. 

Short  methods  of  dividing  by  numbers  that  lack 
but  little  of  a  power  of  ten,  are  given  in  many  text- 
books. They  are  of  little  value  aside  from  the  drill 
afforded  in  mastering  them. 

Example.     8967843-^-98  =  ? 


Form. 


Quotients.       Remainders. 


89678 
1793 

35 


43 
56 
86 
70 


55 


59 

Explanation.  100  is  contained  in  8976843,  89678 
times,  with  a  remainder  of  43.  98  is  contained  89678 
times,  with  a  remainder  of  43,  and  a  further  remain- 
der of  two  for  each  time  100  is  contained.  This 
further  remainder  is  179356.  Proceeding  as  above 
with  this  number  the  first  remainder  is  56  and  the 
further  remainder  is  3586.  Proceeding  in  the  same 


Methods  in  Written  Arithmetic.  45 

way  with  3586,  the  first  remainder  is  86  and  the 
further  remainder  is  70,  a  number  less  than  the 
divisor.  The  sum  of  the  first  remainders  is  255. 
Proceeding  as  above  with  255  the  first  remainder  is 
55  and  the  further  remainder  is  4.  The  sum  of  the 
quotients  is  91508,  and  of  the  remainders  is  59. 


CHAPTER  VI. 


FACTORING. 

IF  pupils  are  to  become  expert  in  arithmetical 
operations  they  must  learn  to  factor  numbers  with 
celerity.  Especially  is  this  the  case  in  Least  Com- 
mon Multiple,  Greatest  Common  Divisor,  Fractions, 
Percentage  and  its  applications,  and  Proportion.  A 
pupil  who  has  had  thorough  drill  in  factoring  will 
perform  nineteen-twentieths  of  the  problems  in  Frac- 
tions, found  in  the  average  written  arithmetic,  with- 
out the  aid  of  his  pencil;  and  few  exercises  do  so 
much  toward  giving  power  to  a  class  as  drill  of  that 
kind. 

As  introductory  to  the  work  the  pupils  should 
memorize  the  factors  of  numbers  to  one  hundred. 
These  should  be,  like  the  multiplication  table,  part 
of  their  stock  in  trade.  They  should  be  fixed  in  the 
memory  so  that  they  will  come  immediately  when 
called  for. 


Methods  in  Written  Arithmetic.  47 

In  testing  a  class  give  dictation  board  exercises, 
with  all  the  pupils  at  the  board,  if  possible,  and  so 
separated  that  they  shall  not  help  each  other.  This 
may  be  done,  as  has  already  been  suggested,  by  let- 
ting them  number,  and  by  giving  one  kind  of  work 
to  the  even  numbers  and  another  to  the  odd. 

The  board  work  should  be  scrupulously  neat.  The 
following  form  is  suggested  : 

28=2X2x7. 
36=2X2X3X3. 

48=2X2  X2  X  2X3. 

The  pupil  may  read  the  above  work  as  an  equa- 
tion, or  as  follows : 

The  prime  factors  of  28  are  two  twos  and  seven; 
of  36  are  two  twos  and  two  threes;  of  48  are  four 
twos  and  three. 

While  it  is  true  that  28=2X2x7,  it  is  not  true 
that  the  prime  factors  of  28  are  2  times  2  times  7, 
as  many  pupils  recite;  much  less  is  it  true  of  "two 
square,  seven." 

In  this  connection  teach  the  definitions  of  Prime, 
Composite,  Factor,  Common  Factor,  Prime  Factor, 
Prime  to  each  other,  Odd,  Even,  Multiple,  Common 


48  Methods  in  Written  Arithmetic. 

Multiple.  The  etymologies  of  these  terms  will  help 
to  a  better  appreciation  cf  their  meaning.  The  una- 
bridged dictionary  will  furnish  them,  and  the  pupils 
should  form  the  habit  of  consulting  its  pages  for  that 
kind  of  aid. 

TESTS    OF   DIVISIBILITY. 

it    All  even  numbers  are  divisible  by  two. 

2.  A  number  is  divisible  by  three  if  the  sum  of 
the  ones  expressed  by  its  digits  is  divisible  by  three. 

3.  A  number  is  divisible  by  four  if  the  number 
expressed  by  the  two  right-hand  digits  is  divisible  by 
four. 

4.  A  number  is  divisible  by  five  if  the  right  hand 
digit  is  zero  or  five. 

5.  A  number  is  divisible  by  six  if  divisible  by  two 
and  three. 

6.  A  number  is  divisible  by  eight  if  the  number 
expressed  by  the  three  right-hand  digits  is  divisible 
by  eight. 

7.  A  number  is  divisible  by  nine  if  the  sum   >f  the 
ones  expressed  by  the  digits  is  divisible  by  nir  j>. 

8.  A  number  is  divisible  by  ten  if  the  right-hand 
figure  is  zero. 


Methods  in  Written  Arithmetic.  49 

9  A  number  is  divisible  by  eleven  if  the  sum  of  the 
ones  expressed  by  the  digits  in  the  odd  orders  equals 
the  sum  of  the  ones  expressed  by  the  digits  in  the 
even  orders,  or  if  the  ditteience  of  these  sums  is  a 
multiple  of  eleven. 

Other  tests  might  be  made  by  combining  these, 
but  enough  have  been  given  to  answer  ordinary  pur- 
poses. 

Pupils  are  troubled  often  in  determining  whether 
a  number  is  prime  or  composite.  They  continue  the 
division  until  they  think  there  is  reasonable  ground 
for  considering  it  prime,  simply  because  the  divisor 
is  large.  They  should  not  be  in  doubt. 

A  number  is  prime  if  successive  primes  have  been 
tried  unsuccessfully  until  the  integral  part  of  the 
quotient  is  bss  than  the  divisor  last  used. 

Since  no  smaller  number  than  the  prime  last  used 
will  divide  it,  no  larger  number  will,  for  if  it  were 
possible  for  a  larger  number  to  divide  it  the  quotient 
would  be  a  smaller  number;  but  the  quotient  is  a 
factor  of  the  dividend,  and  we  have  seen  that  the 
dividend  has  no  factor  smaller  than  the  prime  used 
last. 


50  Methods  in  Written  Arithmetic. 

PRINCIPLES  EMPLOYED  IN  FACTORING. 

1.  A  divisor  of  a  number  is  a  divisor  of  any  multi- 
ple of  that  number. 

This  is  evident  since  every  time  the  number  is  re- 
peated the  divisor  is  repeated. 

2.  A  divisor  of  two  numbers  is  a  divisor  of  their 
sum. 

Each  is  some  number  of  times  the  common  di- 
visor, hence  their  sum  is  some  number  of  times  the 
common  divisor. 

3.  A  divisor  of  two  numbers  is  a  divisor  of  their 
difference. 

Since  each  is  some  number  of  times  the  common 
Hivisor,  if  they  differ  it  must  be  because  one  contains 
the  common  divisor  more  times  than  the  other,  hence 
the  difference  is  some  number  of  times  the  common 
divisor. 

DEMONSTRATION  OF  TESTS. 
TEST  FOR  FOUR. 

i.  Any  number  of  more  than  two  orders  may  be 
regarded  as  some  number  of  hundreds  plus  the  num- 
ber expressed  by  the  two  right-hand  figures. 


Methods  in  Written  Arithmetic.  51 

Since  one  hundred  is  divisible  by  four,  any  num- 
ber of  hundreds  must  be  by  Principle  i.  If  the 
number  expressed  by  the  two  right-hand  figures  is 
divisible  by  four  the  whole  number  is  by  Principle  2. 


TEST   FOR  FIVE. 

2.  Any  number  of  more  than  one  order  may  be 
regarded  as  some  number  of  tens  plus  the  number 
expressed  by  the  right-hand  figure.     If  the   right- 
hand  figure  is  zero  the  number  is  tens,  and  since  one 
ten  is  divisible  by  five  the  number  must  be  divisible 
by  five  by  Principle  2. 

TEST  FOR   EIGHT. 

3.  Any  number  of  more  than  three  orders  may  be 
regarded  as  some  number  of  thousands  plus  the  num- 
ber expressed  by  the  three  right-hand  figures.    Since 
one    thousand    is    divisible  by   eight,   any  number 
of  thousands  is  so  divisible  by  Principle  i.      If  the 
number  expressed    by  the   three    right-hand   digits 
is  divisible  by  eight  the  whole  number  must  be  by 
Principle  2 


52  Methods  in  Written  Arithmetic. 

TEST   FOR   NINE. 

4.  Any  number  may  be  separated  into  two  parts, 
one  of  which  is  a  multiple  of  nine  and  the  other  the 
sum  of  the  ones  expressed  by  the  digits.  Since  by 
definition  the  first  part  is  divisible  by  nine,  it  follows 
that,  if  the  second  part  is,  the  whole  number  is,  by 
Principle  2. 

The  first  statement  in  the  above  demonstration 
may  need  elaborating  somewhat. 

One  ten  is  one  more  than  nine  units;  that  is,  it  is 
one  more  than  a  multiple  of  nine.  Two  tens  are 
two  more  than  a  multiple  of  nine.  Any  number  of 
tens  is  as  many  more  than  a  multiple  of  nine  as  there 
are  tens.  The  same  may  be  said  of  one  hundred, 
two  hundreds,  or  any  number  of  hundreds;  so  of 
thousands,  ten  thousands,  etc. 

Or,  making  a  general  statement,  since,  in  a  deci- 
mal system,  any  unit  equals  ten  of  the  next  lower 
order,  it  must  follow  that  any  decimal  unit  is  one  of 
the  next  lower  order  more  than  a  multiple  of  nine ; 
hence  one  in  any  order  expresses  a  number  that  is 
one  more  than  the  multiple  of  nine;  two,  a  number 
that  is  two  more  than  a  multiple  of  nine;  any 
digit,  a  number  as  many  more  than  a  multiple  of 


Methods  in  Written  Arithmetic.  53 

nine  as  the  number  of  ones  expressed  by  the  digit. 
If  the  sum  of  these  several  remainders  is  a  multiple 
of  nine,  the  number  must  be. 

(Frame  a  demonstration  of  the  test  for  three  from 
the  above.) 

Pupils  should  be  able  to  apply  these  tests  rapidly. 
Statements  similar  to  the  following  will  be  found 
convenient  : 

Factor  25864. 

This  number  is  divisible  by  8  because  864  is  so 
divisible.  Removing  this  factor  the  quotient  is  3233. 
This  number  is  odd,  hence 'is  not  divisible  by  any 
even  number.  Since  the  sum  of  the  ones  expressed 
by  the  digits  is  1 1  the  number  is  not  divisible  by  3 
or  any  of  its  multiples,  because,  etc.  On  trial  I  find 
it  is  not  divisible  by  7.  It  is  not  divisible  by  n, 
because,  etc.,  and  in  the  same  way  the  successive 
primes  are  tried  to  53,  which  is  found  to  give  a  quo- 
tient of  61.  The  prime  factors  then  are  2,  2,  2,  53, 
61. 

If  3233  were  prime,  how  long  should  the  trials 
have  continued?  Why? 

The  explanation  of  the  test  for  divisibility  by 
eleven  will  be  understood  if  the  following  facts  are 
observed : 


54  Methods  in  Written  Arithmetic. 

1.  One  ten  is  one  less  than  a  multiple  of  eleven; 
two  tens  are  two  less,  and  any  number  of  tens  are 
as  many  less  as  there  are  tens. 

A  similar  statement  may  be  made  for  thousands, 
hundred  thousands,  ten  millions,  etc.  But  tens, 
thousands,  hundred  thousands,  ten  millions,  etc.,  oc- 
cupy orders  whose  numbers,  counting  from  the  right, 
are  even.  Hence  a  digit  standing  in  an  order  whose 
number  is  even  expresses  a  number  which  is  as 
many  less  than  a  multiple  of  eleven  as  the  number 
of  ones  expressed  by  the  digit. 

2.  In  a  similar  manner  it  may  be  shown  that  a 
digit  standing  in  an  order  whose  number  is  odd  ex- 
presses a  number  which  is  as  many  more  than  a  mul- 
tiple of  eleven  as  there  are  ones  expressed  by  the 
digit. 

3.  If  these  remainders  are  equal  they  balance  each 
other  and  the  number  is  a  multiple.     If  one  set  of 
remainders   exceeds    the   other    by   a    multiple   of 
eleven  it  must  follow  that  the  whole  number  is  a 
multiple  of  eleven. 

Although  the  tests  of  divisibility  are  indispensable 
aids  to  rapid  work,  it  requires  considerable  drill  to 
render  pupils  expert  in  their  use.  Rapid  dictation 
exercises  can  be  used  advantageously  to  secure  this 


Methods  in  Written  Arithmetic.  55 

result.     Send  all  the  pupils  to  the  board,  or  see  that 
all  are  supplied  with  paper  or  slates. 

AN    EXERCISE    IN   THE   THREE   TEST. 

The  teacher  pronounces   the  following  numbers 
as  rapidly  as  the  pupils  can  work:     2841;   52689; 
48263;  2875356;  796;  87234;  and  so  following. 
The  pupils  write : 

2841,     15,     yes. 
52689,     30,     yes. 
48263,     23,     no. 
2875356,     36,     yes. 
796,     22,     no. 
87234,   -24,     yes. 

The  entire  class  can  be  tested  in  the  time  required 
for  one  by  the  oral  method.  The  exercise  tests  the 
ability  to  write,  add  and  divide.  The  work  should 
be  very  neat,  and  the  teacher  should  see  the  results. 
These  exercises  should  be  continued  until  the 
pupils  can  apply  the  tests  promptly  and  accurately. 


CHAPTER  VII. 


CANCELLATION. 

Cancellation  is  a  method  of  shortening  the  work 
in  problems  involving  only  multiplication  and  di- 
vision. 

PRINCIPLES. 

1.  Dividing  any  one  of  a  series  of  factors  by  any 
number  divides  their  product  by  that  number. 

2.  Dividing  dividend   and   divisor  by  the  same 
number  does  not  change  the  quotient. 

The  first  principle  seems  very  simple,  yet  most 
pupils  trip  in  its  application.  They  do  not  recognize 
24X36X75,  for  example,  as  a  number.  This  they 
should  be  accustomed  to  do, — a  number  that  is 
partially  factored.  If  a  teacher  should  ask  his  class 
how  he  can  divide  such  a  number  by  three,  many 
will  answer  "Divide  each  factor  by  three."  Show  the 
error.  Read  the  number  as  24  times  36  times  75. 

(56) 


Methods  in   Written  Arithmetic.  57 

One  third  of  it  is  8  times  36  times  75,  24  times  12 
times  75,  or  24  times  36  times  25. 

The  effect  of  the  presence  of  o  in  a  list  of  factors 
is  often  misunderstood.  Try  your  classes  on  the  fol- 
lowing: 8X6X0X4=?  Some  will  answer  "48;"  others 
"4;"  fewer  will  answer  "nothing,"  as  they  should. 
Make  it  clear  that  whenever  o  enters  as  a  factor  the 
product  is  o. 

24X45X60 

18X20 

EXPLANATION. 

This  is  a  problem  in  division  in  which  divisor  and 
dividend  are  partially  factored.  I  shorten  the  opera- 
tion by  dividing  divisor  and  dividend  by  9.  One- 
ninth  of  the  divisor  is  two  times  twenty.  One-ninth  of 
the  dividend,  is  twenty-four  times  five  times  sixty.  I 
further  shorten  the  operation  by  dividing  divisor  and 
dividend  by  four.  One-fourth  of  the  dividend  is  six 
times  five  times  sixty,  etc, 

Pupils  will  soon  be  able  to  omit  larger  factors. 
Thus :  one-eighteenth  of  the  divisor  is  twenty.  One- 
eighteenth  of  the  dividend  is  twelve  times  five  times 
sixty. 

When  the  language  work  is  well  advanced  pupils 


58  Methods  in    Written  Arithmetic. 

should  perform  many  problems  without  explanation, 
getting  the  result  in  the  shortest  time  possible.  Do 
not  allow  pupils  to  carry  dead-weight  in  the  shape  of 
common  factors.  Encourage  pupils  to  find  short  pro- 
cesses. 

GREATEST    COMMON    FACTOR. 

A  factor  of  a  number  is  a  divisor  of  the  number,  01 
a  number  that  is  contained  in  it  an  integral  number 
of  times. 

A  common  factor  of  two  or  more  numbers  is  a 
divisor  of  each  of  them. 

The  greatest  common  factor  of  two  or  more  num- 
bers is  the  greatest  number  that  will  divide  each  of 
them. 

A  prime  factor  of  a  number  is  a  divisor  that  is  a 
prime  number. 

Any  number  is  divisible  by  the  product  of  two  or 
more  of  its  prime  factors. 

Pupils  should  learn  to  recognize  all  the  divisors, 
whether  prime  or  composite,  of  numbers. 

EXERCISES. 

1.  Name  all  the  prime  divisors  of  30,  39,  42,  50, 
etc. 

2.  Name  all  the  divisors  of  the  same  numbers. 


Methods  in   Written  Arithmetic.  59 

3.  Name  all  the  common  prime  divisors  of  40  and 
65  ;  of  32  and  48 ;  of  30,  45,  60,  etc. 

4.  Review  (3)  naming  all  the  common  divisors  in 
each  case. 

5.  Review  (4)  naming  the  greatest  common  factor 
in  each  case. 

PROBLEM. 

What  is  the  greatest  common  factor  of  40,  48,  56? 

FORM. 

40^2X2X2X5          ) 
48=2X2X2X2X3    [-2X2X2=8 
56=2x2x2x7          ) 

EXPLANATION. 

Two  is  a  prime  factor  of  each  of  the  numbers, 
hence  a  factor  of  the  greatest  common  factor. 
Another  two  is  a  prime  factor  of  each  of  the  num- 
bers, hence  a  factor  of  the  greatest  common  factor. 
A  third  two  is  a  prime  factor  of  each  of  the  numbers, 
hence  a  factor  of  the  greatest  common  factor. 

Since  there  are  no  other  common  prime  factors  the 
product  of  two,  two  and  two,  or  eight,  is  the  greatest 
common  factor.  When  the  method  of  building 
the  g.  c.  f.  is  mastered  the  explanation  may  be 
thortened. 


60  Methods  in  Written  Arithmetic. 

Give  many  problems  in  which  only  the  result  is 
asked  for  in  order  to  secure  facility. 

Observe  that  the  g.  c.  f.  is  the  product  of  the 
common  prime  factors. 

Now  drill  the  pupils  in  finding  the  g.  c.  f.  by  in- 
spection. This  can  be  done  with  readiness  when  the 
numbers  are  not  large. 

It  is  well  for  the  pupils  to  use  the  words/tf^/w  and 
divisor  interchangeably. 

An  examination  of  the  difference  of  two  numbers 
or  of  the  difference  between  one  of  them  and  some 
number  of  times  the  other,  will  often  disclose  their 
g.  c,  f.  and  save  time. 

What  is  the  g.  c.  f.  of  2373  and  2499?  Their  dif- 
ference is  126;  its  prime  factors  are  2,  3,  3,  7.  Since 
a  common  divisor  of  two  numbers  is  a  divisor  of  their 
difference,  the  g.  c.  f.  of  2373  and  2499  must  divide 
126.  Since  a  common  divisor  of  two  numbers  is  a 
divisor  of  their  sum  the  g.  c.  f.  of  2373  and  126  must 
be  divisor  of  2499  and  must  be  the  g.  c.  f.  of  2373 
and  2499,  hence  I  examine  only  126  and  2373. 

Three  is  a  factor  of  2373,  hence  a  factor  of  the 
g.  c.  f.  Seven  is  a  factor  of  2373,  hence  it  is  also  a 
factor  of  the  g.  c.  f.  Since  these  are  the  only  common 


Methods  in  Written  Arithmetic.  61 

prime  factors  their  product  is  the  g.  c.  f.  of  2373  and 
2499. 

What  is  the  g.  c.  f.  of  387  and  2754?  2754  is  45 
more  than  7  times  387.  Since  a  factor  of  a  number 
is  a  factor  of  any  multiple  of  that  number,  all  the 
factors  of  387  are  factors  of  7  times  387. 

Seven  times  387  and  2754  have  no  common 
prime  factors  that  are  not  also  common  to  387 
and  2754,  hence  by  preceding  principles  the 
g.  c.  f.  of  45  and  387  is  the  g.  c.  f.  of  387 
and  2754. 

If  the  smaller  number  is  a  divisor  of  the  larger 
it  is  the  g.  c.  f.  sought. 

This  method  may  be  used  with  more  than  two 
numbers  by  finding  the  g.  c.  f.  of  the  two,  then  of  that 
number  and  a  third,  etc. 

The  methods  that  have  been  given  of  finding  the 
greatest  common  factor  of  numbers  are  ample  for  all 
cases,  but  the  method  employed  in  algebra  where 
quantities  are  given  whose  factors  are  not  found  by 
the  application  of  the  common  theorems  is  presented 
in  many  arithmetics  and  may  deserve  discussion. 

Its  demonstration  is  too  difficult  for  children  and 


62  Methods  in  Written  Arithmetic. 

should  not  be  required  unless  the  class  is  unusually 
mature. 

Find  the  g.  c.  f.  of  3139  and  4307. 

The  g.  c.  f.  of  these  numbers  cannot  be  greater 
than  3139.  If  3139  will  divide  4307  it  is  their  g.  c.  f. 
It  is  contained  once  with  a  remainder  of  1168,  hence 
it  is  not  their  g.  c.  f.  A  common  divisor  of  two  num- 
bers is  a  divisor  of  their  difference,  hence  the  g.  c.  f. 
of  3139  and  4307  must  divide  1168;  it  consequently 
cannot  be  greater  than  1168.  If  n  68  will  divide 
3139  it  will  also  divide  4307,  the  sum  of  1168  and 
3139,  and  will  be  the  g.  c.  f.  sought.  It  is  contained 
twice  with  a  remainder  of  803,  hence  it  is  not  the  g.  c. 
f.  of  3139  and  4307. 

Since  the  g.  c.  f.  must  divide  1168  it  must  divide 
twice  1 1 68,  or  2336.  (Why?)  Since  it  must  divide 
2336  and  3139  it  must  divide  their  difference,  or  803. 
(Why  ?)  Hence  it  cannot  be  greater  than  803.  If 
803  will  divide  1 1 68  it  will  divide  2336.  (Why?) 
3139,  (Why?)  and  4307,  (Why?)  and  will  be  the 
divisor  sought  It  is  contained  in  1168  once  with  a 
remainder  of  365,  hence  it  is  not  the  g.  c.  f.  sought 
Apply  the  same  method  of  reasoning  to  365. 

It  is  contained  in  803  twice  with  a  remainder 
of  73.  Apply  the  same  method  of  reasoning  to  73. 


Methods  in  Written  Arithmetic.  63 

This  method  depends  upon  the  three  principles  of 
factoring  already  given  and  is  simply  a  substitution 
of  smaller  numbers  which  have  the  same  g.  c.  i".  as 
the  numbers  given. 


CHAPTER  VIII. 


MULTIPLES. 

A  multiple  of  a  number  is  an  integral  number  of 
times  that  number. 

A  common  multiple  of  two  or  more  numbers  is  a 
multiple  of  each  of  them. 

The  least  common  multiple  of  two  or  more  num- 
bers is  the  least  number  that  is  a  multiple  of  each  of 
them. 

What  is  the  1.  c.  m.  of  56,  72,  96. 

FORM. 

56  =  2.   2.   2.  7,  j 

72  =  2.   2.   2.  3.  3,  C  2.  2.  2.  2.  2.3.3.7  =  2016=1.0.  m. 

96  =  2.   2.   2.   2.   2.  3,       ) 

EXPLANATION. 

Since  the  1.  c.  m.  of  these  numbers  must  contain 
96  it  must  contain  its  prime  factors,  which  I   use  as 
factors  of  the  1.  c.  m.     I  wish  to  so  change  this  pro- 
(64) 


Methods  in   Written  Arithmetic.  65 

duct  that  it  shall  also  contain  72.  This  I  do  by  in- 
troducing the  additional  factor  3.  I  now  have  a  pro- 
duct that  will  contain  96  and  72.  I  wish  to  so  change 
this  product  that  it  will  also  contain  56.  This  I  do 
by  introducing  the  additional  factor  7.  The  1.  c.  m. 
is  the  product  of  five  twos,  two  threes  and  seven. 
This  number  contains  such  prime  factors  and  such 
only  as  are  necessary  to  produce  the  several  num- 
bers. 

The  g.  c.  f.  of  these  numbers  is  eight.  The  un- 
common prime  factors  are  2,  2,  3,  3,  7.  The  1.  c.  m. 
of  two  or  more  numbers  is  the  product  of  their 
g.  c.  f.  and  their  uncommon  prime  factors. 

ANOTHER  METHOD. 

When  pupils  have  acquired  considerable  readiness 
in  seeing  the  factors  that  compose  a  number,  the 
work  may  be  abridged  very  materially  by  employing 
what  may  be  called  the  "inspection"  method. 

DIRECTIONS. 

1.  If  any  number  is  a  factor  of  any  other,  strike 
it  out,  since  its  factors  are  contained  in  the  other 
number. 

2.  Take  the  largest  number  and  compare  one  of 


66  Methods  in   Written  Arithmetic. 

the  others  with  it,  retaining  the  factors  of  the  second 
not  found  in  the  first  and  striking  out  the  others. 

3.  Compare  a  third  with  the  first  and  remaining 
factors  of  the  second,  retaining  factors  found  in  nei- 
ther and  striking  out  the  others  and  so  proceed  with 
the  remaining  numbers. 

EXAMPLE. 
Find  the  1.  c.  m.  of  12,  18,  24,  27. 

EXPLANATION. 

Since  12  is  a  factor  of  24,  I  strike  it  out.  8  con- 
tains all  the  prime  factors  of  24  not  found  in  27, 
hence  I  retain  the  8  and  strike  out  the  24.  18  has 
no  factors  not  contained  in  8  or  27,  hence  8  times 
27  is  the  1.  c.  m.  of  the  numbers. 

A  THIRD   METHOD. 

Some  problems  include  numbers  not  easily  factor- 
ed, and  much  time  may  be  required  in  searching  for 
divisors.  The  work  may  be  shortened  by  taking  two 
of  the  numbers,  finding  their  g.  c.  f.  by  either  of  the 
last  two  methods,  dividing  one  by  the  g.  c.  f.  and 
multiplying  the  other  by  this  quotient.  Compare 
the  multiple  thus  obtained  with  the  third  number 
and  so  following.  This  method  depends  upon  the 
principle  already  stated.  The  1.  c.  m.  of  two  num- 


Methods  in  Written  Arithmetic,  (-^ 

bers  is  the  product  of  their  g.  c.  f.  and  the  uncom- 
mon prime  factors,  or,  what  is  the  same  thing,  it  is 
the  product  of  one  of  the  numbers  and  all  the  prime 
factors  of  the  other  not  contained  in  the  first. 

EXAMPLE. 
Find  the  1.  c.  m.  of  4087  and  4757- 

EXPLANATION. 

I  find  that  the  difference  of  these  numbers  is  670; 
its  prime  factors  are  2,  5,  67.  I  see  that  neither  2, 
nor  5  is  a  factor  of  4087.  By  trial  I  find  that  67  is 
contained  in  4087  sixty-one  times;  67  is  therefore 
a  factor  of  4757,  for  a  factor  of  two  numbers  is  a 
factor  of  their  sum;  moreover  67  is  the  only  common 
factor.  61  times  4757  is,  then,  the  1  c,  m.  of  the 
two  numbers. 

Remember  the  old  proverb,  "Practice  makes  per- 
fect." 


CHAPTER  IX. 


FRACTIONS. 

Fractions  contain  little  that  is  new.  If  the  pre- 
ceding work  has  been  mastered,  there  should  be  but 
little  difficulty.  The  new  thing  is  the  method  of 
expressing  the  kind  of  units  that  compose  the  num- 
ber. 

A  Fractional  number  is  a  number  that  is  composed 
of  one  or  more  Fractional  units. 

A  Fractional  unit  is  a  relative  unit  that  is  formed 
by  separating  an  absolute  unit  into  equal  parts. 

The  expression  f  is  read  five  sixths.  This  is  the 
only  case  in  which  6  is  read  sixths.  It  is  this  fact, 
perhaps,  which  causes  most  trouble.  The  numeratoi 
of  a  Fractional  number  is  the  number  of  fractional 
units  that  compose  the  fractional  number. 

The  Denominator  of  a  Fractional  number  is  the 
number  of  the  fractional  units  required  to  make  an 
(68) 


Methods  in  Written  Arithmetic.  69 

absolute  unit.  The  Denominator  may  also  be  de- 
fined as  the  number  of  equal  units  into  which  an 
absolute  unit  has  been  separated.  It  thus  indicates 
the  size  of  the  fractional  units. 

In  the  discussion  of  integers  it  was  stated  that 
there  are  two  things  to  be  considered  in  respect  to 
each  number;  first,  how  many?  second,  what  kind? 
In  integers  the  first  question  is  answered  by  the 
shape  and  order  of  succession  of  the  figures;  the 
second,  by  the  position  of  the  right-hand  figure  with 
respect  to  the  decimal  point.  In  fractional  num- 
bers the  same  condition  exists.  The  numerator  an- 
swers the  first  question,  and  the  denominator  the 
second. 

We  have,  then,  little  or  nothing  that  is  new. 

A  Common  Fraction  is  one  whose  numerator  is 
written  above  and  denominator  below  a  short  hori- 
zontal line. 

A  Decimal  Fraction  is  one  whose  denominator 
expresses  a  power  of  ten. 

The  denominators  of  decimal  fractions  are  not 
usually  written ;  they  are  generally  expressed  by  the 
position  of  the  right-hand  figure  of  the  numerator 
respect  to  the  decimal  point. 


70  Methods  in  Written  Arithmetic^ 

Now  define  Proper  Fraction,  Improper,  Simple, 
Complex,  Compound,  and  give  and  explain  the 
Principles  of  Fractions  given  in  all  text-books. 

As  with  integers,  so  with  fractions,  there  are  but 
two  operations.  They  may  be  united  or  separated. 

The  Reduction  of  a  number  consists  in  changing 
its  denomination  without  altering  its  value.  The 
following  reductions  are  performed  in  the  study  of 
Common  Fractions : 

I.  A  whole   or  mixed   number   to   an   improper 
fraction. 

II.  An  improper  fraction  to  a   whole   or   mixed 
number. 

III.  A  fraction  to  its  lowest  terms. 

IV.  A  fraction  to  an  equivalent  fraction   having 
any  denominator. 

V.  Fractions  to  equivalent  fractions  having  a  com- 
mon denominator. 

I. 

iffi  is  called  a  mixed  number  because  there  are 
two  kinds  of  units,  viz. :  seven  units  of  one  kind  and 
three  of  another.  The  reduction  consists  in  chang- 
ing this  number  to  one  in  which  there  is  but  one 
kind  of  unit. 


Methods  in  Written  Arithmetic.  71 

Since  in  one  there  are  eight  eighths,  in  seven  there 
are  seven  eights  of  eighths,  or  fifty-six  eighths.  Fifty- 
six  eighths  plus  three  eighths  equal  fifty-nine 
eighths.  Form  a  rule. 

II. 

An  improper  fraction  is  a  simple  number;  that  is, 
the  units  are  of  one  kind.  The  reduction  consists 
in  changing  it  to  another  simple  number  (whole),  or 
to  a  mixed  number. 

-2g-7-  =  ?  Since  in  one  there  are  eight  eighths,  in 
27  eighths  there  are  as  many  ones  as  there  are  eights 
in  27.  There  are  y/§  eights  in  27:  hence  27 
eighths=3^  ones.  Form  a  rule. 

III. 

A  fraction  is  in  its  lowest  terms  when  numerator 
and  denominator  are  prime  to  each  other,  that  is, 
when  they  have  no  common  prime  factors. 

Since  the  g.  c.  f.  is  the  product  of  the  common 
prime  factors  it  is  clear  that  if  the  terms  be  divided 
by  their  g.  c.  f.  the  resulting  quotients  will  be  prime 
to  each  other. 

|5.  =  ?  Dividing  the  terms  by  5,  their  g.  c.  f.,  the 
resulting  fraction  equals  |.  It  is  obvious  that  |f  = 


72  Methods  in  Written  Arithmetic. 

|,  for  although  there  are  but  \  as  many  fractional 
units,  each  is  5  times  as  large  as  before ;  or,  consid- 
ering the  fraction  as  a  problem  in  division  we  have 
the  principle  that  dividing  the  divisor  and  dividend 
by  the  same  number  does  not  change  the  quotient. 
Form  a  rule. 


IV. 


Change  ^  to  twenty-firsts. 

Since  in  one  there  are  f }-,  in  •]•  there  are  -£T,  and 
in  f  there  are  three  threes  of  twenty-firsts,  which 
equal  -fa. 

Change  f  to  tenths. 

Since  in  one  there  are  |-g-,  in  J  there  are  *S,  and 
in  |  thert  are  5  times  *2,  which  =  5i.  Form  a  rule. 

ANOTHER   METHOD. 

Since  multiplying  numerator  and  denominator  by 
the  same  number  does  not  change  the  value  of  the 
fraction,  multiply  both  terms  by  such  a  number  as 
will  make  the  given  denominator  equal  the  required 
denominator.  How  shall  this  multiplier  be  found? 
Form  a  rule. 


Methods  in  Written  Arithmetic.  73 

V. 

In  changing  fractions  to  equivalent  fractions  hav- 
ing a  common  denominator,  why  do  we  select  for 
the  required  denominator  a  multiple  of  the  given 
denominators?  Change  f,  f,  ^  to  equivalent 
fractions  having  15  for  a  denominator.  What  kind  of 
fractions  result?  Why  select  the  1.  c.  in.  of  the 
denominators? 

The  "Inspection  method"  of  finding  the  1.  c.  m. 
will  be  found  especially  convenient.  Use  either 
method  given  in  IV  in  making  the  reductions. 

In  making  these  several  reductions  teachers  will 
find  it  advisable  to  employ  object  work,  even  with 
pupils  of  grammar  grades.  It  is  neither  necessary 
nor  expedient  to  use  objects  in  all  cases,  as  the 
pupils  are  mature  enough  to  imagine  them  present. 
Require  a  clear  statement  of  the  exact  process  in  I 
if  the  7^6  were  apples.  Do  the  same  in  each  case, 
III  will  be  found  quite  difficult. 


CHAPTER  IX. 


ADDITION   OF    FRACTIONSo 

As  in  addition  of  integers,  only  like  numbers  can 
be  added. 

Like  fractions  are  those  which  have  the  same  frac- 
tional unit.  Since  the  several  numerators  express 
the  number  of  fractional  units  in  the  respective  frac- 
tions, their  aggregate  expresses  the  sum  of  the  frac- 
tional units  in  all  of  the  fractions,  hence  the  familiar 
rule. 

Unlike  fractions  are  those  which  do  not  have  the 
same  fractional  unit.  /<3  and  ^  cannot  be  united. 
Before  addition  is  possible  they  must  be  made  alike, 
or  changed  to  equivalent  fractions  having  a  common 
denominator.  In  making  this  change  encourage  the 
use  of  the  "inspection1'  method  of  finding  the  1.  c.  m. 
of  the  denominator. 

Do  no  unnecessary  work.  Do  not  change  mixed 
(74) 


Methods  in  Written  Arithmetic.  75 

numbers   to  improper  fractions,   and  thus  unneces- 
sarily burden  all  of  the  subsequent  processes. 

Form  a  rule. 

Encourage  work  like  the  following: 

fH-f-|-f{H-i$=  **  I  w^sn  to  nnd  h°w  many  ones 
there  are  in  the  sum  of  these  fractions.  -J  lacks  \  of 
being  a  unit.  I  make  \  of  ^  thus  leaving  \\  which 
equals  \.  '-3-j-i— i.  £  lacks  \  of  being  a  unit;  I 
make  f  of  T6g-,  leaving  T9F.  |-{-3-=i.  I  am  still  to 
unite  -^g-  and  -|.  -f-  lacks  f-  of  being  a  unit.  Since  I 
cannot  conveniently  make  fifths  of  sixteenths,  I 
change  -j9^  to  something  of  which  fifths  may  be 
formed.  T9g-=  <-<>-;  $$  are  required  to  make  |.  |  + 
1=1-  H — it=8o;  hence,  the  result  equals  3^. 

As  in  reductions,  have  the  pupil  think  the  ^rocess 
with  objects. 

SUBTRACTION    OF    FRACTIONS. 

What  is  the  fundamental  principle  of  subtraction? 
If  minuend  and  subtrahend  are  not  alike  what  must 
be  done  with  them? 

An  effort  should  be  made  to  show  the  'one-ness'  of 
all  subtractions. 


76  Methods  in  Written  Arithmetic. 

Since  &  and  }"  are  unlike,  I  change  them  to  like 
numbers.  How?  f=ff.  li=tf-  Since  I  cannot 
take  £,L  from  f  |,  I  take  one  unit  from  8  units  and 
reduce  it  to  seventy-sevenths,  etc.  Form  a  rule. 

An  explanation  similar  to  the  following  will  be 
found  helpful. 

Separate  \  into  two  parts,  one  of  which  is  |.  Since 
I  cannot  conveniently  make  thirds  of  eighths,  I 
change  \  to  a  fraction  of  which  thirds  may  be  made. 
i=|l.  J£  are  required  to  make  f  .  If  J-{-  be  sepa- 
rated into  two  parts,  one  of  which  is  ££,  the  other 
will  be  25T.  Describe  the  process  as  it  would  be  if 
performed  with  objects. 

MULTIPLICATION    OF   FRACTIONS. 

Review  the  definitions  given  in  multiplication  of 
integers.  If  the  definition  of  multiplier  be  correct 
it  is  obvious  that  it  cannot  be  a  proper  fraction. 


This  problem  may  be  read,  "multiply  j8^-  by  6,"  or, 
"unite  6  eights  of  fifteenths."  6  eights  of  fifteenths 
are  \\  which  =,  etc. 

The  6  groups,  each  containing  j85  are  reduced  to 
one  group  containing  |f.  This  in  turn,  is  separated 


Methods  in  Written  Arithmetic.  77 

into  3  groups,  each  containing  J  §-,  with  a  remainder 
of  fe  or  £,  i.  e.,  it  is  reduced  to  a  mixed  number. 

The  simplest  method  of  multiplying  a  fraction  by 
an  integer,  and  the  one  most  easily  illustrated  by  the 
use  of  objects,  is  the  multiplication  of  the  numerator 
by  the  integer,  since  this  gives  the  result  obtained 
by  uniting  as  many  of  the  numerators  as  are  ex- 
pressed by  the  integer. 

In  such  a  problem  as  -fe  X  5,  if  the  numerator  be 
multiplied  by  5,  f  f  will  result.  This  will  be  reduced 
to  lowest  terms  by  dividing  both  terms  by  5,  hence 
three  operations  will  have  been  performed,  viz. :  the 
multiplication  of  7  by  5,  the  division  of  35  by  5,  and 
the  division  of  15  by  5.  It  is  evident  that  the  first 
two  should  be  omitted,  since  the  second  simply  un- 
does the  first;  hence  in  such  cases  no  opportunity 
of  dividing  the  denominator  should  be  lost. 

A  second  method  of  explaining  the  effect  of  divid- 
ing the  denominator  by  the  integer  in  the  problem 
given  above,  is  as  follows : 

By  dividing  the  denominator  of  j75  by  5  a  frac- 
tional number  results,  each  of  whose  units  is  equal 
to  5  of  the  former  units.  Since  the  number  of  units 


78  Methods  in  Written  Arithmetic. 

is  the  same  in  the  two  cases  it  is  obvious  that  the 
second  number  equals  five  of  the  first. 

In  the  problem  -j^-x6,  there  is  an  opportunity  to 
divide  the  denominator  by  a  factor  of  the  integer. 

The  formal  statement  is  substantially  as  follows :  I 
first  multiply  fs  by  3  by  dividing  its  denominator  by 
3,  giving  -|,  Multiplying  f  by  2  the  result  is  J^. 
The  more  common  form  of  expression  is :  Omitting 
the  common  factor  3,  etc. 

The  following  is  a  rule  for  multiplying  a  fraction 
by  an  integer: 

a.  Divide  the  denominator  of  the  fraction  by  the 
integer  if  possible,  or,  if  not  possible, 

b.  Divide  the  denominator  of  the  fraction  by  a 
factor  of  the  integer,  if  possible,  and  multiply  the  nu- 
merator by  the  remaining  factor,  or,  if  not  possible, 

c.  Multiply  the  numerator  of  the  fraction  by  the 
integer. 

With  the  definition  of  multiplication  here  employed 
such  a  problem  as  multiply  8  by  |  is  absurd.  The 
problem  should  be  read,  find  f  of  8.  There  is  in 
such  a  problem  a  division  (second  case),  and  a  mul- 
tiplication. The  multiplicand  is  \  of  8,  and  the 
multiplier  is  5.  Give  an  analysis  and  form  a  rule. 


Methods  in  Written  Arithmetic.  79 

Such  expressions  as  -jj-X  \  should  be  read  as  \  of  |. 
The  habit  of  omitting  common  factors  from  numer- 
ator and  denominator  should  be  established. 

Since  such  problems  as  T75  x  ^  =  ?  involve  both 
multiplication  and  partition,  a  full  analysis  of  the 
process  is  deferred  until  Division  has  been  discussed. 


CHAPTER  X. 


DIVISION   AND    PARTITION    OF    FRACTIONS. 

Review  the  definitions  of  Division  and  Partition, 
The  cases  arising  in  this  discussion  are  illustrated 
in  the  following  problems : 

(i)    8+!  =  ?  (2)    H-f=? 

(3)     fH=?  (4)     il-8=? 

(i)  This  problem  is  usually  read,  "Divide  8  by 
J."  The  meaning  evidently  is  "Separate  8  into  equal 
parts,  each  of  which  is  -jj,"  or,  "Find  how  many  threes 
of  fifths  there  are  in  8." 

In  order  that  8  may  be  separated  into  numbers, 
each  of  which  is  |,  I  change  8  to  fifths.  B=-^°-.  In 
-45°-  there  are  13^  threes  of  fifths. 

This  is  the  plan  that  would  generally  be  followed 
if  objects  were  employed  to  illustrate  the  problem. 
(What  shorter  plan  with  the  objects?) 
(80) 


Methods  in  Written  Arithmetic.  81 

A  rule  based  on  the  above  would  be  as  follows: 

To  divide  an  integer  by  a  fraction,  reduce  the  in- 
teger to  the  same  denomination  as  the  fraction,  and 
divide  the  numerator  of  the  dividend  by  the  numer- 
ator of  the  divisor. 

(2)  |-~|=?  How  many  twos  of  sevenths  are 
there  in  £?  This  may  also  be  read,  "Separate  f  into 
twos  of  sevenths." 

In  order  that  \  may  be  separated  into  two  of  sev- 
enths, I  change  |  to  a  fraction  of  which  sevenths 
may  be  made.  -|=r||-.  ±%  are  needed  to  form  f.  In 
||  there  are  2|  tens  of  thirty-fifths. 

Hence,  |H--|  =  2*. 

'     9  I  B 

Describe  the  process  if  objects  were  employed. 

Form  a  rule  similar  to  (i). 

A  short  explanation  of  Division  by  a  fraction  in- 
volves the  following : 

To  divide  one  by  a  fraction,  change  one  to  the 
denomination  of  the  fraction  and  divide  the  numer- 
ator of  the  dividend  by  the  numerator  of  the  divisor. 
In  such  a  case,  however,  the  numerator  of  the  divi- 
dend will  be  the  same  as  the  denominator  of  the 
divisor.  Hence,  to  divide  one  by  a  fraction,  divide 
the  denominator  of  the  divisor  by  its  numerator. 


82  Methods  in  Written  Arithmetic. 

The  following  is  a  brief  explanation  of  (2). 

H-? 

Were  the  dividend  i  the  quotient  would  be  \. 
Since  the  dividend  is  £  of  i  the  quotient  is  \  of  J. 

fH=? 

Thoughtful  pupils  will  observe  that  expressions  of 
this  character  are  not  problems  in  division  as  we 
have  used  the  term.  Neither  do  they  fall  under 
Partition,  as  will  shortly  appear.  They  involve 
Comparison.  The  question  really  is,  how  far  will  4 
go  toward  making  |?  or,  f  is  what  part  of -£?  %=l$j- 

4—18.      J_ Ois  .10     or     5     Of  2. 8 
5  —  35'      35    lb    28»   U1    14    U1    3  IT' 

They  may,  however,  be  regarded  as  problems  in 
division,  and  may  be  analyzed  according  to  the 
model  given  for  division  by  a  fraction. 

(4)  As  division  has  been  defined  it  is  obviously 
impossible  to  divide  £f  by  8,  since  ^f  cannot  be 
separated  into  numbers,  each  of  which  is  8.  This  is 
a  problem  in  Partition.  £f  may  be  separated  into  8 
equal  parts,  ^f  -=-8  should  be  read,  find  ^  of  £|,  or. 
separate  \%  into  8  equal  parts. 

Pupils  should  see  what  "Striking  out"  common 
factors  means. 


Methods  in  Written  Arithmetic.  83 

This  has  been  shown  in  the  multiplication  of  a 
fraction  by  an  integer;  let  us  observe  the  effect  in 
problems  like  (4). 


Observing  the  factor  7  in  14  and  21,  the  problem 
may  be  read:  Find  £  of  -f  of  -if.    -|  of  -f|=^-.    ^  of 


Why  does  multiplying  the  denominator  of  a  frac- 
tion divide  the  fraction? 

The  denominator  of  a  fraction  shows  the  size  of 
the  fractional  units  which  compose  the  fraction. 
This  it  does  by  showing  the  number  of  fractional 
units  into  which  the  absolute  unit  has  been  sepa- 
rated. If  this  number  be  increased  the  size  of  the 
fractional  unit  will  be  correspondingly  diminished. 

ANOTHER  VIEW. 


Since  ^  of  5,  mixed  number,  I  separate  each 
eighth  into  three  equal  parts,  and  have  -|-f  .  ^  of  ^-f 
is  2\.  The  operation  involves  the  multiplication  of 
the  numerator  and  denominator  by  three,  and  the 
division  of  the  resulting  numerator  by  three  —  three 


84  Methods  in  Written  Arithmetic. 

operations,  two  of  which  may  be  omitted,  since  the 
second  undoes  the  first. 

We  are  now  ready  to  analyze  such  problems  as 
^X?f=? 

What  is  -H  of  B?    A°ftf-*ofioftf.    i<* 


The  short  statement  for  the  same  is,  omit  equal 
factors  from  numerator  and  denominator,  etc. 

It  is  worth  while,  however,  to  have  a  pupil  know 
whether  he  is  multiplying  or  dividing  when  he  omits 
factors. 

COMPLEX   FRACTIONS. 

A  complex  fraction  is  usually  defined  as  "a  fraction 
which  has  a  fraction  in  one  or  both  of  its  terms." 
Are  all  so-called  complex  fractions  really  fractions? 
A  fraction  has  been  defined  as  an  expression  for  one 
or  more  of  the  equal  parts  of  a  unit.  The  complex 
fraction  whose  numerator  is  three-and-a-half,  and 
whose  demominator  is  four,  evidently  falls  under  the 
definition,  but  what  shall  we  say  of  those  whose  de- 
nominators are  fractions?  Can  we  conceive  of  a 
unit  as  being  divided  into  %  equal  parts?  Evidently 
the  idea  is  absurd. 


Methods  in  Written  Arithmetic.  85 

If  the  numerator  is  equal  to  or  greater  than  one, 
and  the  denominator  is  integral,  the  expression  is  a 
fraction.  If  the  numerator  is  less  than  one,  or  if 
the  denominator  contains  a  fraction,  the  expression  is 
simply  a  problem  in  division — fractional  only  in  form. 

It  seems  wiser,  at  first,  to  regard  them  simply  as 
problems  in  divison,  and  to  have  them  so  explained. 
Pupils  should  be  able  to  read  them  as  fractions,  how- 
ever, thus:  6^4 — \%  may  be  read  as  "the  complex 
fraction  whose  numerator  is  6%  and  denominator 
^."  It  maybe  read  also  as  a  problem  in  division: 

6%-TV 

As  problems  in  division  they  present  nothing  new. 
It  is  worth  while,  however,  for  pupils  to  understand 
that  the  laws  of  simple  fractions  are  equally  appli- 
cable to  them;  that  is,  numerator  and  denominator 
may  be  multiplied  and  divided  by  the  same  number 
without  changing  the  value  of  the  fraction,  and  when 
proficiency  has  been  acquired  in  reducing  them  by 
the  ordinary  method,  let  them  try  the  following : 

(*•)  3/^ — 4l/2'  Since  %  is  a  common  factor  of 
the  terms,  divide  each  term  by  it.  This  problem 
illustrates  a  large  class. 

(2.)  J/% — T52.  Multiply  both  terms  by  24,  the 
1.  c.  m.  of  the  denominators.  Multiplying  the  nu- 


86  Methods  in  Written  Arithmetic. 

merator  first  by  8,  the  product  is  seven ;  multiplying 
7  by  3,  the  product  is  21. 

Multiplying  the  denominator  first  by  12,  the  pro- 
duct is  5 ;  multiplying  this  product  by  2  (the  remain- 
ing factor  of  24),  the  product  is  10. 

Any  complex  fraction  may  be  simplified  very 
readily  by  this  process.  If  one  or  both  of  the  terms 
are  mixed  numbers,  do  not  reduce  them  to  improper 
fractions.  Reduce  4.1 — 5^.  I  multiply  both  terms 
by  10,  the  1.  c.  m.  of  their  denominations.  5  times 
the  numerator  is  21;  2  times  21=42,  etc. 

It  should  be  remembered  that  there  is  only  one 
way  of  making  pupils  quick  and  accurate  workers ; 
they  must  perform  problems  by  the  hundred.  Send 
the  class  to  the  board  so  that  their  methods  of  work 
can  be  seen,  then  dictate  problems  and  require  rapid 
work.  Permit  no  erasing.  Let  every  mark  remain. 
If  errors  are  made,  and  discovered  by  the  pupils,  the 
previous  work  can  be  crossed  and  corrections  made, 
but  all  that  has  been  done  should  appear  upon  the 
board.  If  pupils  realize  that  their  errors  are  to  face 
them,  they  will  be  more  careful  about  making  them. 
In  a  twenty-minute  recitation  each  pupil  can  per- 
form from  ten  to  thirty  such  reductions  as  are  given 
above. 


CHAPTER  XI. 


THE    METHODS    OF     FINDING    THE    PART   WHICH    ONE 
NUMBER    IS    OF   ANOTHER. 

The  two  questions  A  is  how  many  times  B  ?  and 
A  is  what  part  of  B  ?  are  in  substance  the  same. 
The  question  usually  assumes  the  first  form  if  A  is 
greater  than  B;  if  A  is  less  than  B  the  question 
usually  assumes  the  second  form.  The  answer  to 
the  second  is  always  fractional,  while  the  answer  to 
the  first  may  be  either  integral  or  mixed. 

FORMS   OF   PROBLEMS- 

1.  7  is  what  part  of  10? 

One  is  ^  of  10,  hence  7  must  be  ^  of  10.  (How 
many  objects  are  needed  to  illustrate  it?) 

2.  ^  is  what  part  of  4? 

Since  only  like  numbers  can  be  compared,  I  charge 
4  to  thirds.     4=  V-     The  question  now  is,  f  is  what 
part  of !/?    §  are  ^  of  y.     Show  with  objects. 
(87) 


88  Methods  in  Written  Arithmetic. 

3.    f  is  what  part  of  |? 

Since  only  like  numbers  can  be  compared,  I  make 
the  fractions  alike.  ^=fa;  |  — If-  The  question 
now  is,  etc. 

It  will  be  seen  that  the  same  result  is  obtained  in 
each  case  if  the  first  number  be  divided  by  the  sec- 
ond ;  hence  the  common  rule  : 

'Divide   the  number  expressing  the  part  by  the 
number  of  which  it  is  a  part. 

These  relations  may  be  shown  very  easily  by  fold- 
ing paper  squares. 

To  illustrate  (3),  fold  the  square  into  sixths,  and 
then  to  twenty-fourths.  Show  to  how  many  of  the 
twenty-fourths  the  |  and  f  are  severally  equal,  and 
then  how  far  the  latter  will  go  toward  making  the 
former. 

STRAIGHT   LINE   ANALYSIS. 

Familiarity  with  this  topic  will  enable  pupils  to 
perform  most  problems  involving  only  multiplication 
and  division  in  a  small  part  of  the  time  usually 
required. 

Most  of  the  problems  in  multiplication  and  divi- 
sion of  fractions,  many  in  the  applications  of  per- 
centage, almost  all  in  simple  and  compound  proper 


Methods  in  Written  Arithmetic.  89 

tion,  and  very  many  others  admit  of  such  a  solution. 
Indeed,  since  Proportion  belongs  more  properly  to 
algebra,  it  is  questionable  whether  we  should  attempt 
it  in  arithmetic,  unless  the  class  has  abundant  time 
or  is  quite  mature. 

DIRECTIONS. 

1.  Begin  with  a  number  which  is  of  the  denomi- 
nation of  the  required  result.     Since  the  product  is 
like  the  multiplicand,  and  since  the  divisions  may  all 
be  of  the  second  case,  in  which  the  quotient  is  like 
the  dividend,  each  result  will  be  like  the  number 
with  which  we  begin. 

2.  Fractions  should  appear  in  neither  term  and 
should  be  put  upon  the  straight  line  a  term  at  a 
time. 

If  the  number  with  which  we  begin  is  mixed, 
change  it  to  an  improper  fraction. 

PROBLEM. 

If  27  men  in  18  days  of  10  hours  each  dig  a  ditch 
1 80  rods  long,  6  feet  wide,  and  3  feet  deep,  of  5 
degrees  of  hardness,  in  how  many  days  of  nine  hours 
each  will  45  men  dig  a  ditch  300  rods  long,  8  feet 
wide  and  4%  feet  deep,  of  7^3  degrees  of  hardness? 


90  Methods  in  Written  Arithmetic. 

ANALYSIS. 

Since  the  question  asks  for  a  number  of  days  we 
begin  with  18  days.  Drawing  a  horizontal  line,  "18 
days"  is  written  above  the  left  end. 

One  man  could  perform  the  work  in  27  times  18 
days,  which  is  expressed  by  writing  27  as  a  factor 
of  the  dividend.  We  now  have  an  expression  for  the 
number  of  days  required  for  one  man  to  do  the  work 
if  27  men  could  do  it  in  18  days.  Since  45  men  are 
to  do  the  work,  they  can  accomplish  it  in  one-forty- 
fifth  of  this  number  of  days,  which  is  expressed  by 
writing  45  as  a  factor  of  the  divisor.  We  now  have  an 
expression  for  the  number  of  days  required  for  45 
men  to  do  the  work  if  28  men  can  do  it  in  18  days. 

If  these  men  worked  only  one  hour  a  day,  ten 
times  as  many  days  would  be  needed.  This  is  ex- 
pressed Dy  writing  10  as  a  factor  of  the  dividend. 
What  ia  now  expressed? 

Since  the  45  men  are  to  work  9  hours  a  day  only 
one  ninth  as  many  days  are  required,  which  is  ex- 
pressed by  writing  9  as  a  factor  of  the  divisor.  What 
is  now  expressed  ? 

The  ditch  is  180  rods  long.  Were  it  but  one  rod 
long,  only  one  one-hundred-eightieth  as  many  days 


Methods  in   Written  Arithmetic.  gi 

would  be  needed.  This  is  expressed  by  writing  180 
as  a  factor  of  the  divisor. 

Since  the  proposed  ditch  is  to  be  300  rods  long,  300 
times  as  many  days  are  necessary,  which  is  expressed 
by  writing  300  as  a  factor  of  the  dividend.  (The  pu- 
pil should  be  able  to  tell  what  is  expressed  at  any 
step.)  Were  the  ditch  but  one  foot  wide,  45  men 
could  dig  it  in  one- sixth  this  number  of  days, — ex- 
pressed, etc.  Since  the  ditch  is  to  be  eight  ft.  wide, 
eight  times  this  number  of  days  are  needed, — ex 
pressed,  etc. 

Were  the  ditch  but  one  foot  deep,  one-third  this 
number  of  days  would  be  required, — expressed,  etc. 
Were  the  ditch  only  one-half  a  foot  deep,  only  one- 
half  of  this  number  of  days  would  be  necessary. 
Since  the  ditch  is  to  be  nine-halves  feet  deep,  nine 
times  this  number  of  days  are  needed,  and  so  on 
through  the  solution.  When  the  work  is  indicated  it 
will  assume  the  following  form : 

18x27  x 


45X9Xi8ox6x3X2XSX3 

All  common  factors  should  now  be  omitted. 
The  work  is  necessarily  slow  at  first,  but  the  teach- 
er should  persist  until  it  is  understood  and  can  be 


92  Methods  in  Written  Arithmetic. 

given  with  accuracy  and  rapidity.  It  is  a  perfect 
machine,  but,  unlike  most  machine  processes,  it  can- 
not be  used  unless  it  is  understood. 

The  problems  usually  given  in  Cancellation  are 
simple  enough  to  introduce  the  topic.  When  they  arc 
exhausted  turn  to  Simple  Proportion.  These  will  be 
found  to  be  more  difficult.  Follow  these  with  problems 
from  Compound  Proportion,  similar  to  the  one  given 
above. 

While  introducing  the  work,  those  who  find 
it  very  difficult  should  be  aided  by  questions,  although 
they  should  not  be  too  suggestive. 

Emphasize  the  fact  that  in  analysis  "we  go  arouna 
by  the  unit;11  that  is,  if  we  are  told  that  19  men  cai\ 
do  a  piece  of  work  in  24  days,  and  are  asked  how 
many  days  will  be  required  for  50  men  to  do  the  same 
work,  we  first  find  the  number  of  days  that  one  man 
will  require  to  do  the  work. 


CHAPTER  XII. 


DECIMAL   FRACTIONS. 

A.  decimal  fraction  is  one  whose  denominator  is  a 
a  power  of  ten.  The  denominator  instead  of  being 
written,  as  in  the  common  fraction,  is  usually  express- 
ed by  the  position  of  the  right-hand  figure  of  the 
numerator  with  respect  to  the  decimal  point.  In  an 
early  number  of  this  series  directions  were  given  for 
writing  them.  These  should  be  reviewed  with  care. 
The  pupils  should  learn  the  number  of  any  order  at 
the  right  of  the  point  so  that  it  can  be  given  as  soon 
as  the  name  is  pronounced. 

Frequent  dictation  exercises  will  be  found  useful 
in  fixing  these  facts.  Require  statements  similar  to 
the  following: 

"I  write  629  millionths  by  making  9  stand  in  the 
sixth  order  at  the  right  of  the  decimal  point."  In  or- 
der to  write  readily  the  pupil  must  see: 


94  Methods  in   Written  Arithmetic. 

(i.)  How  many  figures  are  necessary  to  express 
the  numerator. 

(2.)  Where  the  right-hand  figure  must  stand  to 
make  the  number  express  the  required  denomination, 
and 

(3.)  How  many  zeros  (if  any)  must  precede  the 
numerator. 

REDUCTIONS. 

1.  To  change  a  decimal  fraction  to  a  common  frac- 
tion erase  the  decimal  point,  write  the  denominatcr, 
and  reduce  to  lowest  terms. 

This  exercise  will  afford  an  excellent  opportunity 
to  review  complex  fractions. 

2.  To  change  a  common  fraction  to  a  decimal. 
(0)  Divide  the  numerator  by  the  denominator,  or 
(<J)  Do  anything  to  the  fraction,  that   you  have  a 

right  to  do,  that  will  make  its  denominator  a  power 
of  ten;  then  erase  the  denominator  and  express  it  by 
the  position  of  the  decimal  point. 

The  first  of  these  methods  should  be  very  easy  to 
the  pupil.  As  has  been  said,  a  fraction  may  be  re- 
garded as  a  problem  in  division.  Seven-sixteenths 
is  one  sixteenth  of  seven. 

In  simple  division,  when  the  first  term  of  the  div- 


Methods  in  Written  Arithmetic.  95 

idend  would  not  contain  the  divisor,  it  was  reduced 
to  the  next  lower  denomination.  Do  the  same  with 
these  problems. 

ANALYSIS. 

Change  seven-sixteenths  to  a  decimal  fraction. 

7  =  7.0.  (Read  70  tenths.)  1-16  of  7.0 =.4  with 
a  remainder  of  .6.  .6=. 60.  1-16  of  .60=. 03  with  a 
remainder  of  .12.  .12=. 120.  1-16  of  .120=. 007 
with  a  remainder  of  .008.  .008=. 0080.  1-16  of 
.0080  =  . 0005.  Hence,  7-16=. 4375. 

Oblige  the  pupils  to  keep  the  decimal  point  in 
position  all  of  the  time  until  the  reduction  is  thor- 
oughly understood.  Do  not  allow  them  to  say  "I 
annex  zeros  to  the  numerator  and  divide  it  by  the 
denominator,  etc." 

Only  those  common  fractions  which  in  their  lowest 
term  have  in  the  denominators  the  factors  2  or  5, 
or  both,  and  no  others,  can  be  changed  to  "pure 
decimals." 

The  reason  for  this  becomes  clear  when  it  is  ob- 
served that  with  each  reduction  of  the  numerator 
only  the  factors  2  and  5  are  introduced. 

The  second  method  is  much  briefer  in  many  cases, 
but  requires  more  skill  in  its  use.  Complex  fractions 


96  Methods  in  Written  Arithmetic. 

whose  denominators  are  aliquot  parts  of  some  power 
of  ten  may  be  reduced  at  once  by  multiplying  both 
terms  by  the  quotient  arising  from  dividing  the 
power  of  ten  by  the  "aliquot." 

EXAMPLES. 

(i).  yfij.  Multiply  both  terms  by  8. 
(2).  gVv  Multiply  both  terms  by  3. 
(3).  -g1^.  Multiply  both  terms  by  if 

The  number  by  which  to  multiply  the  terms  of  a 
common  fraction  may  be  told  by  inspecting  the 
denominator. 

Change  fa  to  a  decimal  fraction. 

The  denominator  is  the  product  of  four  twos.  To 
change  this  fraction  to  a  decimal  fraction  the  denom- 
inator must  be  made  a  power  of  ten.  Since  ten  is 
the  product  of  two  and  five,  any  power  of  ten  is  the 
product  of  an  equal  number  of  twos  and  fives;  hence 
multiply  both  terms  of  the  fraction  by  the  fourth 
power  of  five. 

If,  when  the  fraction  is  in  its  lowest  terms,  other 
prime  factors  than  two  and  five  are  found  in  the 
denominator,  such  factors  must  be  paired  with  mixed 
numbers  or  fractions  to  produce  a  power  of  ten ;  hence 


Methods  in  Written  Arithmetic.  97 

the  numerator  will  be  a  mixed  number  and  the  frac- 
tion will  not  be  a  "pure"  decimal. 

ADDITION   AND    SUBTRACTION. 

There  is  nothing  new  in  addition  and  subtraction. 
The  methods  and  explanations  previously  employed 
are  applicable  and  should  be  used. 

MULTIPLICATION. 

Multiplication  of  decimals  presents  nothing  new 
except  the  "pointing"  of  the  product. 

The  explanations  given  in  multiplication  of  com- 
mon fractions  need  no  modification  for  application 
to  "decimals." 

Such  expressions  as  8.62  X- 023  =  ?  should  be  read 
.02 j  of  8.62  =  ? 

ANALYSIS. 

One-thousandth  of  8.62  =.00862,  obtained  by 
moving  the  decimal  point  three  places  to  the  left. 
.023  of  8.62=123  times  00862. 

Most  arithmetics  explain  the  philosophy  of  "point- 
ing" by  changing  the  "decimals"  to  common  frac- 
tions, performing  the  multiplication  and  then  chang- 
ing the  product  back  to  the  original  form. 


98  Methods  in  Written  Arithmetic. 

It  may  also  be  explained  as  follows : 

When  the  multiplier  is  units  (of  the  first  order), 
the  product  is  like  the  multiplicand  (has  as  many 
"places").  When  the  multiplier  is  of  a  denomination 
lower  than  units,  the  product  is  the  same  part  of  the 
product  obtained  by  considering  the  multiplier  units 
that  the  given  multiplier  is  of  the  multiplier  so  con- 
sidered. 

DIVISION. 

In  division  of  decimals  the  forms  of  explanation 
in  common  fractions  may  be  employed. 

In  all  such  problems  the  divisor  will  be  an  integer 
or  a  "decimal."  If  it  is  an  integer  the  problem  falls 
under  partition  and  should  be  so  explained.  If  it  is 
a  "decimal"  it  will  be  of  the  same  denomination  as 
the  dividend,  or  of  a  higher  or  lower  denomination. 

When  of  the  same  denomination  the  analysis  is 
identical  with  that  first  given  for  the  division  of  a 
fraction  by  a  fraction.  When  of  a  lower  denomina- 
tion the  dividend  may  be  reduced  to  the  same  de- 
nomination and  the  same  analysis  be  employed. 
When  of  a  higher  denomination  it  may  be  reduced 
to  the  same  denomination  as  the  dividend  and  com- 
pared with  it;  or, 


Methods  in  Written  Arithmetic.  99 

In  all  such  cases  divisor  and  dividend  may  be  mul- 
tiplied by  a  number  that  will  make  the  divisor  inte- 
gral. The  explanation  may  then  be  given  by  division 
or  partition,  as  the  case  may  be.  If  the  divisor  is 
less  than  the  dividend,  it  may  be  division;  if  greater, 
it  will  be  partition. 

The  "pointing"  may  be  explained  briefly  as  follows : 

Since  the  dividend  corresponds  to  the  product, 
and  the  divisor  and  quotient  to  the  factors,  it  must 
follow  that  the  quotient  must  have  a  number  of 
"decimal  places"  which  added  to  the  number  in  the 
divisor  will  equal  the  number  in  the  dividend;  or, 

If  the  divisor  were  units  (of  the  first  order)  the 
quotient  would  be  like  the  dividend.  See  Partition. 
If  the  divisor  is  of  a  denomination  lower  than  units, 
the  quotient  is  as  many  times  the  quotient  obtained 
by  considering  the  divisor  units  as  the  divisor  so 
considered  is  times  the  given  divisor. 

Require  the  pupils  to  explain  the  problem  care- 
fully, considering  the  divisor  units.  He  will  see  that 
the  quotient  is  like  the  dividend.  In  making  the 
correction  for  the  given  divisor  he  will  see  that  the 
period  is  moved  as  many  places  to  the  right  in  the 
quotient  as  there  are  figures  at  the  right  of  the  point 
in  the  divisor. 


CHAPTER  XIII. 


COMPOUND    NUMBERS. 

This  subject  contains  more  information  thai  may 
be  made  interesting  than  any  other  topic  in  the  ordi- 
nary arithmetic. 

Too  often  the  only  attention  it  receives  is  the  me- 
chanical learning  of  tables,  definitions  and  rules,  and 
the  more  mechanical  grinding  out  of  results  in  per- 
forming problems.  The  definitions  and  tables  must 
be  learned,  and  there  is  no  escape  from  a  certain 
amount  of  drudgery;  but  much  can  be  done  to  relieve 
the  work  of  its  most  irksome  feature. 

A  compound  number  is  a  concrete  number  which  is 
composed  of  units  of  more  than  one  denomination, 
but  is  reducible  to  a  simple  number. 

Or,  a  compound  number  is  a  concrete  number  in 

which  the  scale  is  not  ten. 

(100) 


Methods  in  Written  Arithmetic. 


The  latter  definition,  of  course,  excludes  Federal 
money. 

Number  has  been  defined  as  the  measure  of  the 
relation  between  quantities. 

Compound  numbers  are  measures. 


They  measure  . 


i.  Money. 


2.    Extension. 


a.  In  one  direction 

> 

or  length. 

b.  In  two  directions 

or  surfaces. 

c.  In   three  direc- 

tions, or  vol- 
umes. 


3.  Weight. 

4.  Time. 

5.  Angles. 

6.  Miscellaneous. 


Measuring  is  the  process  of  finding  how  many 
times  a  quantity  contains  another  quantity,  which 
we  call  the  unit  of  measure. 

The  result  is  a  number  and  simply  gives  us  definite 
id«as  as  to  the  amount  of  the  thing  measured, 
or,  in  other  words,  enables  us  to  understand  each 


loa  Methods  in  Written  Arithmetic. 

The  first  requisite -for  measuring  is  a  unit. 

MONEY  is  usually  defined  as  a  medium  of  exchange; 
that  is,  it  is  that  instrument  by  means  of  which  the 
various  commodities  of  commerce  pass  from  hand  to 
hand.  It  furnishes  a  standard  of  value;  it  is  a 
measure. 

The  exchange  of  one  article  of  use  or  consumption 
for  another  is  called  barter.  Pupils  will  see  how  in- 
convenient this  method  of  exchange  might  be.  A  has 
a  certain  article,  but  prefers  to  have  another.  He  must 
find  a  person  who  has  what  A  desires,  and  who,  also, 
prefers  what  A  has.  This  method  prevails  among 
rude  peoples,  but  the  advantages  of  a  money  system 
soon  become  apparent  to  an  advancing  race, 
hence,  when  a  nation  reaches  a  certain  grade  of  civ- 
ilization, some  kind  of  a  medium  of  exchange  is  adopt- 
ed, or  in  other  words,  money  appears. 

Teachers  will  find  it  profitable,  in  this  connection, 
to  read  the  article  on  "Money"  in  some  standard 
cyclopedia. 

A  scale  is  the  statement  of  the  number  of  units 
in  each  order  required  to  make  one  of  the  next 
higher. 


Methods  in  Written  Arithmetic.  103 


FEDERAL  MONEY. 

What  is  the  scale  in  U.  S.  money  ?  The  unit  is  the 
dollar.  What  is  the  origin  of  the  word  dollar!  Ex- 
plain the  origin  of  the  dollar-sign.  What  is  a  dime  ? 
What  is  the  origin  of  the  word  ?  Answer  the  same 
questions  for  cent  and  mill. 

What  coins  are  issued  by  the  government  ?  What 
is  the  building  called  in  which  they  are  made  ?  What 
it  the  weight  of  a  silver  dollar  ?  What  part  of  the 
coin  is  alloy  ?  Why  is  alloy  used  ?  What  metal  is 
employed?  Silver  coins  weigh  how  many  times  as 
much  as  gold  coins  of  equal  value  ?  How  much  is 
an  ounce  of  gold  worth  ? 

A  cubic  foot  of  gold  weighs  about  1200  pounds 
avoirdupois ;  how  much  is  it  worth  ? 

How  many  kinds  of  paper  money  are  there?  What 
is  the  name  of  each  kind  ?  About  how  much  of  each 
kind  is  thsre  ?  If  all  the  money  in  this  country  wen 
divided  equally  among  the  inhabitants  about  how 
much  would  each  have  ? 

Teachers  must  determine  how  much  of  this  kind  of 
work  will  be  profitable. 


104  Methods  in  Written  Arithmetic* 

REDUCTION. 

Reduction  is  the  process  of  changing  the  denomi- 
nation of  a  number  without  changing  its  value. 

There  are  two  forms  of  reduction.  Reduction 
Ascending  is  the  process  of  changing  a  number  to  a 
number  of  the  same  value,  but  of  a  higher  denomi- 
nation. 

In  the  same  way  define  Reduction  Descending. 

Do  not  talk  about  changing  a  number  from  a 
lower  to  a  higher  denomination.  Omit  the  words  in 
Italics. 

FORM   OF  ANALYSIS. 

Change  428  cents  to  mills. 

1.  Since  in  i  cent  there  are  10  mills,  in  428  cents 

there  are  428  tens  of  mills,  or  4280  mills. 

It  will  be  seen  that  this  method  makes  the  larger 
number  the  multiplier,  and  is  not  always  the  most 
convenient. 

2.  Were  there  only  i  mill  in  a  cent,  in  428  cents 
there  would  be  428  mills.     Since  there  are  10  mills 
in  a  cent,  in  428  cents  there  are  ten  428*3  of  mills, 
or  4280  mills. 


Methods  in  Written  Arithmetic,  105 

3.  Since  there  are  10  mills  in  a  cent,  in  any  num- 
ber of  cents  there  are  10  mills  for  every  cent;  hence 
in  428  cents  there  are  ten  428*5  of  mills. 

One  form  of  analysis  is  sufficient  for  an  ordinary 
class.  If  only  one  is  used,  the  third  will  be  found 
most  convenient. 

REDUCTION   ASCENDING. 

Change  4820  mills  to  cents. 

1.  Since  in  one  cent  there  are  10  mills,  in  4820 
mills  there  are  as  many  cents  as  there  are  tens  in 
4820.     There  are  482  tens  in  4820;  hence  in  4820 
mills  there  are  482  cents, 

2.  If  there  were  only  one  mill  in  a  cent,  in  4820 
mills  there  would  be  4820  cents.    Since  there  are  10 
mills  in  a  cent,  in  4820  mills  there  are  one-tenth  of 
4820  cents,  or  482  cents. 

3.  Since  there  are  10  mills  in  a  cent,  in  any  num- 
ber of  mills  there  are  one-tenth  as  many  cents;  hence 
in  4820  mills  there  are  one-tenth  of  4820  cents,  or 
482  cents. 

These  forms  of  analysis  apply  to  all  problems  in 
Reduction  of  Compound  Numbers. 


io6  Methods  in  Written  Arithmetic. 

There  is  no  reason  why  a  knowledge  of  the  form, 
of  bills  and  accounts  should  not  be  acquired  at  this 
point.  A  few  cents  invested  in  journal  paper  will 
supply  the  needed  material.  Pupils  have  a  fondness 
for  anything  that  looks  like  business.  The  farmer 
boys  should  be  able  to  put  their  fathers'  accounts 
upon  paper  in  a  neat,  accurate  form.  The  exercise 
may  be  substituted  for  the  regular  work  in  penman- 
ship occasionally,  .may  be  assigned  as  a  home  lesson, 
or  may  be  prepared  during  some  school  hour.  Insist 
upon  neatness  and  accuracy.  Many  teachers  are  so 
anxious  about  the  quantity  of  their  work  done  that 
the  quality  is  miserable.  A  little  work  thoroughly 
done  is  infinitely  better  than  three  times  as  much 
done  carelessly. 


ENGLISH   MONEY. 

Pupils  should  become  accustomed  to  recite  the 
scale,  in  any  case,  without  the  table  or  with  it,  and 
forward  and  backward.  Thus,  in  English  money, 
the  scale  is  4,  12,  20." 

The  unit  is  the  'pound  sterling.     A  pound  weight 


Methods  in  Written  Arithmetic.  107 

of  silver  was  anciently  divided  into  240  equal  parts, 
called  pence,  hence  the  origin  of  the  first  part  of  the 
name.  "These  pence  were  called  esterling,  whence 
the  name  'sterling.'  This  is  supposed  by  some 
writers  to  have  been  derived  originally  from  Easter- 
lings,  the  popular  name  of  traders  from  the  Baltic 
and  from  Germany,  who  visited  London  in  the  mid- 
dle ages,  and  some  of  whom  were  probably  employed 
in  coining.  By  others  it  is  supposed  to  be  a  dimin- 
utive of  star,  and  in  some  old  writings  it  is  written 
starling,  the  penny  being  so  called  from  the  star 
often  stamped  upon  it." — Am.  Cyclopedia. 

The  pound  sterling  is  worth  $4.8665  in  U.  S. 
money.  It  is  represented  by  the  coin  called  the 
sovereign  and  by  the  one-pound  bank  note. 

What  is  the  meaning  of  farthing?  Why  so-called? 
Pupils  should  learn  the  names  of  English  coins,  and 
the  value  of  each.  Which  is  purer,  the  English  or 
U.  S.  coin?  A  given  weight  of  English  coin  is 
worth  more  or  less  than  an  equal  weight  of  U.  S. 
coin?  Why? 

Teachers  must  use  their  own  discretion  respecting 
the  number  of  national  currencies  they  will  require 


io8  Methods  in  Written  Arithmetic. 

their  pupils  to  study.  All  will  require  a  mastery  of 
English  money,  and  if  time  permits,  French  and 
German  should  be  added. 


CHAPTER  XIV. 


MEASURES  OF  LENGTH. 

The  unit  of  length  is  the  yard.  It  was  determined 
in  Great  Britain  about  1760. 

Few  pupils  have  any  idea  of  the  care  exercised  in 
.ixing  this  standard.  Experiments  were  made  at 
London  with  pendulums  of  different  lengths  until 
one  was  found  that  beat  seconds,  that  is,  that  vibrat- 
ed 86,400  times  in  an  average  solar  day.  This  pen- 
dulum was  enclosed  in  a  vacuum  to  protect  it  from 
atmospheric  currents,  and  was  suspended  at  the  sea- 
level  in  order  that  it  might  be  as  near  the  earth's 
center  of  gravity  as  possible  in  that  latitude  without 
going  below  the  earth's  surface.  The  length  of  this 
pendulum  was  separated  into  391,393  equal  parts, 
and  360,000  of  them  were  taken  for  a  yard.  A  foot  is 
(109) 


i  jo  Methods  in   Written  Arithmetic, 

one-third  of  this  standard,  and  an  inch  (from  a  word 
meaning  a  twelfth}  is  a  twelfth  of  a  foot. 

Many  pupils  will  be  interested  in  ascertaining  the 
origin  of  the  names  of  the  different  units.  An  un- 
abridged dictionaiy  will  furnish  the  information  need- 
ed in  nearly  all  cases. 

Pupils  should  thoroughly  master  the  tables  of  long 
measure  found  in  their  arithmetics.  Recite  the  scales 
without  the  names  forward  and  backward. 

In  the  reductions  use  the  forms  of  analysis  already 
given. 

MEASURES   OF   SURFACES. 

A  unit  of  length  having  been  selected,  the  surface 
units  are  easily  determined.  Define  surface,  plane, 
rectangle,  square. 

The  area  of  a  surface  is  the  number  which  ex- 
presses the  number  of  times  the  surface  contains  the 
unit. 

What  is  a  square  foot  ?  A  square  rod  ?  There  may 
be  a  difference  between  "a  foot  square"  and  "a  square 
foot,"  but  it  can  be  a  difference  of  shape  only.  The 
fact  that  many  pupils  who  have  finished  the  subject 
are  bothered  by  such  questions  as :  "What  is  the 


Methods  in  Written  Arithmetic.  irr 

difference  between  three  feet  square  and  three  square 
feet?"  shows  that  their  ideas  are  not  clear.  The  sur- 
faces should  be  represented  to  the  eye  until  the  mat- 
ter is  thoroughly  understood. 

It  is  not  the  puipose  to  discuss  the  whole  subject 
of  Compound  Numbers,  but  to  touch  those  points 
with  which  pupils  have  most  trouble. 

A  field  is  200  rods  long  and  120  rods  wide;  what 
is  its  area? 

Most  arithmetics  tell  us  to  multiply  the  length  by 
the  breadth.  A  scene  like  the  following  is  not  un- 
usual : 

Teacher.  What  kind  of  a  number  is  the  multi- 
plier ? 

Class.  The  multiplier  is  always  abstract. 

T.  What  kind  of  a  number  is  the  product? 

C.  The  product  is  always  like  the  multiplicand. 

T.  What  is  the  product  of  8  feet  and  12  feet  ? 

C.  (Unanimously}  96  square  feet. 

T.  And  how  is  it  that  you  have  a  concrete  multi- 
plier and  a  product  unlike  the  multiplicand? 

This  question  is  usually  followed  by  a  period  of 
profound  silence,  as  pupils  begin  to  realize  how  little 
real  faith  they  have  in  principles  which  they  recite 
with  absolute  accuracy. 


U2  Methods  in  Written  Arithmetic. 

Multiplying  the  length  by  the  breadth  never  gave 
the  area  and  never  will.  If  the  number  of  units  in 
the  length  be  considered  an  abstract  number  and  be 
multiplied  by  the  number  of  units  in  the  breadth,  also 
considered  as  an  abstract  number,  the  product  will 
be  an  abstract  number  containing  as  many  units  as 
there  are  surface-units  in  the  area.  It  is  needless  to 
say  that  the  point  is  too  fine ;  it  isn't.  One  statement 
is  false  and  the  other  true.  Teachers  who  permit 
such  propositions  to  pass  unquestioned  will  accept 
equations  like  the  following:  4X2=8+4-^-6—2. 

An  analysis  something  like  the  following  should  be 
required : 

If  the  field  were  a  rod  long  and  a  rod  wide,  it 
would  contain  one  square  rod.  If  it  were  200  rods 
long  and  i  rod  wide,  it  would  contain  200  sq.  rods. 
Since  it  is  200  rods  long  and  120  rods  wide,  it 
must  contain  120  times  2^0  sq.  rods,  or,  etc. 

Show  the  relations  of  the  tables  of  surface  meas- 
ure to  the  table  of  linear  measure.  In  the  reduc- 
tion use  the  form  of  analysis  given. 

MEASURES   OF   VOLUME. 

Define  solid,  cube,  cubic  inch,  cubic  foot,  cubic 
yard,  etc. 


Methods  in  Written  Arithmetic.  113 

The  unit  used  in  measuring  volumes  is  a  cube 
Vvhose  edge  is  some  linear  unit,  or  some  unit  con- 
taining a  specified  number  of  these  cubes. 

Show  the  relations  of  the  tables  of  "solid  meas- 
ure" to  those  of  "long"  and  "square"  measure. 

ANALYSIS. 

What  is  the  volume  of  a  rectangular  parallelepiped 
whose  length  is  8  feet,  width  4  feet,  and  thickness  3 
feet? 

If  the  figure  were  i  foot  long,  i  foot  wide,  and  i 
foot  thick,  it  would  contain  i  cubic  foot.  If  it  were 
8  feet  long,  i  foot  wide,  and  i  foot  thick,  it  would 
contain  8  cubic  feet.  If  it  were  8  feet  long,  4  feet 
wide,  and  i  foot  thick,  it  would  contain  4x8  cubic 
feet=32  cubic  feet.  Since  it  is  8  feet  long,  4  feet 
wide,  and  3  feet  thick,  it  contains  3  X32  cubic  feet= 
96  cubic  feet. 

LUMBER   MEASURE. 

A  foot  of  lumber  contains  144  cubic  inches.  A 
board  that  is  one  inch  thick  and  one  inch  wide,  then, 
must  contain  one-twelfth  of  a  foot  of  lumber  for  each 
foot  of  length.  Twelve-foot  boards  contain  how 
much  lumber  for  each  inch  of  width?  Answer  the 


H4  Methods  in  Written  Arithmetic. 

same  question  for  boards  of  any  length.  Ordinary 
studs  are  2  by  4  inches.  How  much  lumber  do  they 
contain  for  each  foot  of  length? 

Teachers  who  can  borrow  a  lumber  dealer's  meas- 
ure will  find  it  an  object  of  interest  to  their  pupils. 
It  is  furnished  with  several  scales  so  graduated  that 
the  number  of  divisions  which  measure  the  width  of 
a  board  expresses  the  number  of  feet  in  the  board  if 
it  is  one  inch  thick.  What  part  of  an  inch  should 
each  division  be  for  16  feet  lumber?  Answer  the 
same  question  for  boards  of  any  length. 

LIQUID   AND    DRY   MEASURE. 

Pupils  should  see  clearly  that  the  gallon  and  bush- 
el are  measures  of  volume  that  contain  a  specified 
number  of  cubic  inches.  In  liquid  measure  the  gal- 
lon of  231  cubic  inches  in  the  unit,  while  in  dry 
measure  the  Winchester  bushel  of  2150.42  cubic 
inches  is  the  standard. 

Find  the  number  of  cubic  inches  in  the  liquid 
quart,  also  the  number  in  the  dry  quart.  Why  do 
they  differ?  What  is  the  difference  between  the 
liquid  gallon  and  the  dry  gallon?  Why  do  they 
differ  ? 
-  The  apothecaries'  fluid  measure  is  often  omitted, 


Methods  in  Written  Arithmetic.  irs 

yet  the  terms  are  in  common  use.  Pupils  should 
know  what  is  meant  by  a  two-ounce  vial.  But  little 
time  will  be  needed  to  master  this  measure,  and  bot- 
tles of  various  sizes  should  be  brought  to  the  recita- 
tion. Interest  in  the  whole  subject  will  be  greatly 
increased  by  having  the  various  measures  at  hand  for 
the  pupils  to  use. 

The  reductions  are  explained  by  the  methods  of 
analysis  already  given. 

MEASURES   OF  WEIGHT. 

It  should  be  seen  that  all  measures  of  extension 
thus  far  discussed,  have  been  derived  from  the  unit 
of  length— the  yard. 

It  should  now  be  seen  that  all  the  measures  of 
weight  are  derived  from  the  same  source. 

The  unit  from  which  the  various  units  of  troy, 
apothecaries'  and  avoirdupois  weight  are  derived  is 
the  "  troy  pound  of  the  mint."  But  this  unit  is  sim- 
ply a  piece  of  metal  that  weighs  as  much  as  about 
22.8  cubic  inches  of  pure  water  at  its  greatest  densi- 
ty. The  grain  is  one-five  thousand  seven  hundred 
sixtieth  of  the  troy  pound.  7000  of  these  grains  equal 
the  pound  avoirdupois. 


n6  Methods  in   Written  Arithmetic. 

TIME  MEASURE. 

The  year  of  the  calendar  is  the  time  required  for 
the  earth  to  pass  from  a  point  in  its  orbit,  called  the 
vernal  equinox,  to  the  same  point  again. 

Since  this  point  has  a  slight  motion  backward  the 
earth  makes  a  little  less  than  a  complete  revolution 
about  the  sun  each  year.  The  astronomers  call  this 
the  tropical  year;  its  length  is  365  d.,  5  h.,  48  min., 
46.05  sec. 

The  word  day  has  several  meanings.  The  time 
required  for  the  earth  to  turn  once  upon  its  axis  is 
called  a  sidereal^  or  star,  day.  The  time  elapsing  be- 
tween two  successive  passages  of  the  sun  across  the 
same  meridian  is  called  a  solar  day;  it  is  a  little  lon- 
ger than  a  sidereal  day  and  is  of  variable  length. 
The  average  solar  dav  is  the  day  referred  to  in  the 
table  of  time  measure.  It  is  divided  into  24  equal 
parts  called  hours. 

The  ancients  were  unable  to  determine  the  num- 
ber of  days  in  a  year  hence  much  confusion  resulted. 
46  B.  C.,  Julius  Caesar  reformed  the  calendar,  calling 
the  year  365^  days,  the  common  year  being  365 
days,  and  every  fourth  year  366  days.  The  extra  day 
was  introduced  by  repeating  the  24th  of  February. 
As  this  day  was  then  the  sixth  before  the  first  day  of 


Methods  in   Written  Arithmetic.  117 

March  the  years  in  which  it  was  doubled  were  called 
bissextile  years,  or  years  with  two  sixths. 

Since  each  year  lacked  about  eleven  minutes  of 
being  365^  days  long,  the  addition  of  one  day  in 
four  years  was  too  much  by  about  45  minutes. 
Twenty-five  such  additions  in  a  century  would 
make  an  error  of  about  three- fourths  of  a  day. 
1582  the  error  had  amounted  to  about  ten  d? 

Pope  Gregory  corrected  this  error  by  ordering  that 
the  5th  of  October  of  that  year  should  be  called  the 
1 5th.  A  hundred  and  seventy  years  passed  before 
this  reform  was  introduced  into  England  and  her 
colonies.  The  error  had  then  amounted  to  eleven 
days,  consequently  dates  before  1752  are  occasionally 
called  Old  Style,  or  O.  S. 

The  Gregorian  calendar  is  not  perfect,  but  the  er- 
ror is  very  slight  as  the  extra  day  is  added  only  to 
those  years 

(i.)  Whose  numbers  are  divisible  by  4  and  not  by 
400,  and 

(2.)  Whose  numbers  are  divisible  by  400  and  not 
by  4000. 

The  origin  of  the  names  of  the  months  and  of  the 
days  of  the  week  can  be  found  by  consulting  the 
dictionary.  Curious  pupils  will  want  to  know  about 
them. 


n8  Methods  in  Written  Arithmetic. 

For  measurement  of  angles  and  miscellaneous  ob- 
jects consult  the  usual  text -books. 


LONGITUDE   AND    TIME. 

When  the  sun  is  on  the  meridian  of  any  place,  it 
is  noon  at  that  place. 

(The  equation  of  time  is  disregarded  for  obvious 
reasons).  The  time  elapsing  from  one  noon  to  the 
next  is  divided  into  twenty-four  equal  parts,  each  of 
which  is  called  an  hour.  Clocks  and  watches  are 
machines  so  constructed  that  they  revolve  indexes 
over  a  graduated  circle  called  a  dial,  on  which  their 
motions  are  registered.  Timepieces  are  usually 
regulated  to  show  "sun  time."  When  a  watch  marks 
twelve  o'clock  in  the  day  we  expect  to  find  the  sun 
on  the  meridian  of  the  place  at  which  the  watch 
is  set;  hence  one  should  be  able  to  see  by  his 
watch  how  far  the  sun  is  from  the  meridian  at  any 
time. 

Longitude  is  measured  upon  the  circumference  of 
the  equator  or  of  a  parallel,  east  or  west  from  an 
established  meridian.  These  circles  are  perpendicular 
to  the  axis  of  the  earth,  and,  since  the  earth  revolves 
upon  its  axis,  the  circumference  of  each  of  these 


Methods  in   Written  Arithmetic.  119 

circles  revolves  under  the  sun  from  noon  to  noon. 
Each  circumference  is  divided  into  360  equal  parts 
called  degrees,  hence 

An  arc  of  fifteen  degrees  passes  under  the  sun 
each  hour,  an  arc  of  fifteen  minutes  each  minute,  and 
of  fifteen  seconds  each  second.  We  may  then  make 
the  following  table : 


120 


Methods  in  Written  Arithmetic 


5.       3 

•°      g      S 


»       : 


Methods  in   Written  Arithmetic.  121 

The  tables  found  in  some  arithmetics  tell  us  that 
"15  degrees  of  Ion.  equal  an  hour  of  time,  etc." 

Such  statements  are  obviously  incorrect  and  mis- 
leading. 

A  is  in  Ion.  50  degrees  24  min.  east  of  Washing- 
ton, and  B  is  Ion.  17  degrees  12  minutes  west  of 
Washington;  what  is  their  difference  of  Ion.,  or,  in 
other  words,  A  is  how  far  east  of  B,  or  B  is  how  far 
west  of  A? 

It  seems  strange  that  pupils  should  be  bothered 
by  so  simple  a  problem,  yet  experience  proves  that 
they  are.  The  difficulty  arises,  probably,  from  in- 
correct notions,  or  none  at  all,  of  what  longitude  is. 
If  familiar  units  of  measure,  as  the  mile,  the  foot,  etc., 
were  used,  there  would  be  no  trouble.  The  matter 
may  be  made  perfectly  clear  by  requiring  pupils  to 
find  the  distance  between  towns  on  opposite  sides  of 
them.  If  A  and  B  are  on  the  same  side  of  the  prime 
meridian,  a  similar  illustration  should  be  used. 

The  difference  of  longitude  of  A  and  B  is  77  de- 
grees and  36  minutes;  what  is  the  difference  between 
the  local  times  of  the  two  places  ? 

Since  a  difference  of  15  degrees  in  the  longitudes 
of  two  places  makes  a  difference  of  one  hour  in  their 


122  Methods  in  Written  Arithmetic. 

times,  a  difference  of  77  degrees  makes  a  difference 
of  as  many  hours  as  there  are  fifteens  in  seventy- 
seven;  there  are  five  fifteens  in  seventy-seven,  with 
a  remainder  of  two;  hence,  5  hours  and  two  degrees 
remaining;  2  degrees=i2o  minutes;  120  minutes 
plus  36  minutes=i56  minutes,  etc. 

The  difference  in  the  times  of  two  places  is  2 
hours  24  minutes;  what  is  their  difference  of  longi- 
tude? 

Since  a  difference  of  one  minute  in  the  times  of 
two  places  shows  a  difference  of  15  minutes  in  their 
Ion.,  a  difference  of  any  number  of  minutes  of  time 
shows  a  difference  of  15  times  as  many  minutes  of 
Ion.  A  difference  of  24  minutes  of  time,  therefore, 
shows  a  difference  of  15  times  24  minutes  of  Ion., 
or  360  minutes  of  Ion.,  etc.  (Since  pupils  are  inclined 
to  confuse  the  two  kinds  of  minutes,  the  symbols 
should  be  employed  instead  of  the  words). 

Since  a  difference  of  15  degrees  of  Ion.  makes  a 
difference  of  one  hour  in  time,  a  difference  of  one 
degree  in  Ion.  makes  a  difference  of  4  minutes  in 
time,  a  difference  of  i  minute  in  Ion.  makes  a  differ- 
ence of  4  seconds  in  time,  and  a  difference  of  i 
second  in  Ion.  makes  a  difference  of  four-fifteenths 
of  a  second  in  time. 


Methods  in  Written  Arithmetic.  £23 

These  facts  being  mastered,  work  may  often  be 
ehortened. 

ILLUSTRATIONS. 

1.  The  difference   in  Ion.  of  two  places  being 
1 8°,  25',  30",  what  is  their  difference  in  time? 

Since  a  difference  of  15  degrees  of  Ion.  makes  a 
difference  of  an  hour  in  time,  a  difference  of  18 
degrees  of  Ion.  makes  a  difference  of  an  hour  and  12 
minutes  in  time.  Since  a  difference  of  15  minutes 
in  Ion.  makes  a  difference  of  one  minute  in  time,  a 
difference  of  25  minutes  in  Ion.  makes  a  difference 
of  i  minute  and  40  seconds  in  time.  12  min.-)-i 
min.  =  13  min.  Since  a  difference  of  15  seconds  of 
Ion.  makes  a  difference  of  i  second  in  time,  a  differ- 
ence of  30  seconds  of  Ion.  makes  a  difference  of  2 
seconds  in  time;  40  sec.-f-2  sec.=42  sec.;  hence, 
etc. 

2.  The  difference  in  the  times  of  two  places  is  6 
hours,  26  min.,  25  sec.;  what  is  their  difference  in 
ion.? 

(Fill  out  the  following :) 

A  difference  of  6  hours  shows  a  difference  of  90 
degrees.  A  difference  of  26  minutes  shows  a  differ- 


124  Methods  in  Written  Arithmetic, 

ence  of  6  degrees,  with  a  remainder  of  2  minutes. 
A  difference  of  2  minutes  of  time  shows  a  difference 
of  30  minutes  of  longitude,  etc. 

The  matter  should  now  be  cleared  up  by  numer- 
ous illustrations  similar  to  the  following: 

Suppose  that  you  should  start  from  Chicago,  your 
watch  indicating  Chicago  time,  and  after  traveling 
for  a  time  you  should  find  that  the  time  at  some 
station  which  you  are  passing  is  ten  o'clock  A.  M. 
while  your  watch  says  nine  o'clock  A.  M.  Which 
way  have  you  gone?  How  do  you  know?  How 
far  east  or  west  have  you  traveled  ?  How  do  you 
know? 

Mariners  employ  these  calculations  in  determining 
their  longitude  at  sea.  They  are  supplied  with 
watches  called  chronometers,  that  run  with  remark- 
able accuracy.  These  watches  indicate  the  time 
of  the  port  of  departure,  or  of  some  observatory,  as 
Greenwich.  Whenever  a  mariner  looks  at  his  chro- 
nometer, then,  he  sees  Greenwich  time.  By  means 
of  his  compass  he  can  establish  a  north  and  south 
line.  If  the  sky  is  clear,  so  that  he  can  see  the  sun, 
he  can  tell  when  it  is  noon  where  he  is  as  a  farmer 
does — by  noticing  when  his  shadow  makes  a  north 


Methods  in    Written  Arithmetic.  125 

and  south  line.  He  knows  that  it  is  then  noon. 
His  chronometer  tells  him  what  time  it  is  at  his  port 
of  departure.  He  is  thus  enabled  to  determine  his 
longitude.  By  similar  observations,  he  can  also 
ascertain  his  latitude.  The  inability  to  make  these 
calculations,  accounts  for  many  disasters  at  sea — 
"The  weather  had  been  so  cloudy  that  the  captain 
had  been  unable  for  several  days  to  take  an  observa- 
tion." 


CHAPTER  XV. 


PERCENTAGE. 

Percentage  is  that  part  of  Arithmetic  which  treats 
of  computations  by  hundredths. 

One  per  cent,  of  anything  is  one  hundredth  of  it ; 
two  per  cent,  is  two  hundredths  of  it,  etc. 

Per  cent,  may  be  expressed  in  four  ways : 

1.  As  a  decimal  fraction. 

2.  As  a  common  fraction. 

3.  By  the  use  of  the  symbol  of  per  cent. 

4.  By  the  words  percent. 

Pupils  should  be  required  to  express  in  these  four 
ways  any  per  cent,  whether  integral,  fractional  or 
mixed. 

Percentage  presents  three  general  problems,  or 
cases,  and  no  more. 

(126) 


Methods  in  Written  Arithmetic.  127 

First  General  Problem :  To  find   any  per  cent,  of 

any  number. 

Second  General  Problem :  To  find  what  per  cent, 
one  number  is  of  another. 

Third  General  Problem :  To  find  a  number  when 
some  per  cent,  of  it  is  given. 

For  each  of  these  General  Problems  there  are  two 
solutions.  Pupils  should  master  both,  and  should  let 
the  character  of  the  particular  problem  determine 
the  one  to  be  used.  Each  solution  also  involves  cer- 
tain subordinate  problems  that  will  appear  in  the 
discussion. 

THE   FIRST  GENERAL  PROBLEM. 

First  Method.  Find  one  per  cent,  of  the  number 
and  multiply  the  result  by  the  number  expressing 
the  per  cent. 

The  Subordinate  Problem :  To  find  one  per  cent, 
of  a  number.  With  integers  this  is  done  by  moving 
the  decimal  point  two  places  to  the  left ;  with  frac- 
tions, by  dividing  by  100  in  the  most  convenient 
method  possible  in  the  given  case. 

Second  Method.  Change  the  per  cent,  to  a  common 


128  Methods  in   Written  Arithmetic. 

fraction  in  its  lowest  terms,  and  take  such  a  part  of 
the  number  as  the  fraction  expresses. 

Subordinate  Problem :  To  change  any  per  cent,  to 
a  common  fraction.  Erase  the  sign  of  per  cent,  and 
write  100  as  a  denominator. 

If  the  number  expressing  the  per  cent,  is  fractional 
or  mixed,  this  reduction  will  present  a  complex  frac- 
tion. Pupils  should  become  very  expert  in  reducing 
these  fractions  to  simple  ones.  A  knowledge  of  the 
aliquot  parts  spoken  of  in  an  early  article  will  be 
found  very  convenient  in  this  connection. 

ILLUSTRATIVE   PROBLEMS. 

i.  What  is  23  per  cent,  of  864  ? 

One  per  cent,  of  864  is  8.64,  obtained  by  moving 
the  decimal  point  two  places  to  the  left.  23  per  cent, 
of  864  is  23  times  8.64,  etc. 

What  is  13}^  per  cent,  of  690? 

T-ZYi  per  cent,  equals  13^  hundredths,  or  40-300, 
or  2-15.  2-15  of  690=92. 

THE    SECOND    GENERAL   PROBLEM. 

First  Method.    Find  one  per  cent,  of  the  second 


Methods  in  Written  Arithmetic.  129 

number,  and  divide  the  first  number  by  it.  There  ar« 
two  subordinate  problems  in  this  solution: 

(tf.)  Finding  one  per  cent,  of  a  number. 

(^.)  Dividing  by  a  decimal  fraction. 

ILLUSTRATIVE   PROBLEMS. 

i.  1 8  is  what  per  cent. of  72? 

One  per  cent,  of  72  is  .72,  obtained  by  moving  the 
decimal  point  two  places  to  the  left.  18  is  as  many 
per  cent,  of  72  as  18  is  times  .72.  18=18.00.  18.00 
is  25  times  .72, hence  18  is  25  percent,  of  72. 

It  will  be  seen  that  the  second  subordinate  prob- 
lem is  performed  by  reducing  the  dividend  to  the  de- 
nomination of  the  divisor— a  convenient  plan  in  di- 
vision of  decimals. 

This  solution  is  very  simple  and  can  be  per- 
formed rapidly.  The  pupils  should  perform  a  large 
number  of  problems,  and  the  practice  work  should 
be  continued  until  great  facility  is  acquired.  When 
the  problem  is  given  the  first  number  should  be  writ- 
ten at  the  right  of  the  second,  with  a  curved  line  be- 
tween them.  One  hundredth  of  the  divisor  should 
then  be  obtained,  and  the  dividend  should  be  reduc- 
ed to  hundredths.  Insist  that  the  decimal  point  shall 


130  Methods  in   Written  Arithmetic. 

be  in  place.  Do  not  permit  pupils  to  say  "  I  annex 
two  /eroes. "  It  should  be  clear  to  the  pupil  that  the 
dividend  is  reduced  to  hundredths.  When  this  work 
is  completed  the  problem  presents  the  following 
form:  .72)  18.00  (  The  quotient  is,  of  course, 
units. 

If  the  numbers  are  fractional,  the  solution  is  the 
same.  Thus :  3-5  is  what  per  cent,  of  ^  ?  One  per 
cent,  of  #j  is  7-800.  3-5  is  as  many  per  cent,  of  ^j 
as  3-5  is  times  7-800.  Inverting  the  divisor,  the  prob- 
lem assumes  the  form  3  5X800-7.  Canceling  the 
common  factor  five,  the  result  is  480-7,  or  68  4-7, 
hence  68  4-7  per  cent. 

Second  Method.  Find  what  part  the  first  number  is 
of  the  second  and  change  this  part  to  hundredths.  It 
will  be  seen  that  there  are  two  subordinate  problems 
in  this  solution: 

(a).     Finding  what  part  one  number  is  of  another. 

(£).     Changing  a  common  fraction  to  hundredths. 

These,  like  the  preceding  subordinate  problems, 
were  fully  discussed  before,  consequently  should 
not  be  new.  If  they  are  forgotten  review  them. 

ILLUSTRATIVE    PROBLEMS. 

i.  7  is  what  per  cent,  of  15? 


Methods  in  Written  Arithmetic.  131 

7  is  7-fifteenths  of  15.  7-fifteenths=equals  one- 
fifteenth  of  7.00,  or  .46^3  or  46^  per  cent.  The 
quotient  in  this  case  is  hundredths. 

If  the  numbers  are  fractional  the  solution  is  the 
same. 

2.  Four-fifteenths  is  what  per  cent.  of.  seven- 
twelfths  ? 

Four-fifteenths  is  sixteen-thirty-fifths  of  seven- 
twelfths.  Change  this  fraction  to  hundredths  as  be- 
fore. 

The  work  in  percentage  is  a  constant  review  of 
common  fractions.  Little  can  be  done  unless  that 
subject  has  been  mastered. 

The  successful  teacher  sees  far  in  advance  of  his 
class.  He  knows  that  the  lesson  of  to-day  is  a  pre- 
paration for  the  work  of  months  hence,  and  so  he 
builds  against  the  time  to  come. 

THE   THIRD    GENERAL   PROBLEM. 

As  in  the  preceding,  so  in  this  there  are  two 
methods  of  solution. 

ILLUSTRATIVE    PROBLEMS. 

*•  35  ^  7  Per  cent,  of  what  number? 

Since  35  is  7   per  cent,  of  some  number,  one  pei 


132  Methods  in  Written  Arithmetic. 

cent,  of  that  number  is  one-seventh  of  35,  or  5. 
100  per  cent,  of  the  required  number  is  100  times  5, 
or  500. 

2.  36  is  16  per  cent,  of  what  number? 

16  per  cent.  =  sixteen  hundredths,  or  four  twenty- 
fifths.  Since  36  is  four  twenty  fifths  of  some  num- 
ber, one  twenty-fifth  of  that  number  is  one-fourth  of 
36,  etc. 

Let  the  pupils  state  the  two  methods  of  solution  ol 
this  General  Problem. 

Percentage  presents  no  other  problems  than  these. 
When  a  problem  is  presented  the  pupil  must  deter- 
mine where  it  belongs,  if  he  can  locate  it  the  solu- 
tion will  not  be  difficult. 

$4.80  is  33^  per  cent,  more  than  what  number? 

Problems  like  the  above  are  often  assigned  to  a 
Fourth  case;  they  are  simply  forms  of  the  third 
General  Problem.  A  fuller  statement  of  the  pro- 
blem is: 

$4.80  is  33  YI  per  cent,  of  some  number  more  than 
that  number;  what  is  the  number? 

If  $4.80  is  33^  per  cent,  of  some  number  more 
than  the  number,  it  is  100  per  cent,  of  the  numbet  + 
33J^i  per  cent,  of  the  number,  or  133^  percent  of 
the  number. 


Methods  in   Written  Arithmetic.  133 

•28  is  20  per  cent,  less  than  what  ?  means  28  is 
20  per  cent,  of  some  number  less  than  the  number; 
what  is  the  number  ? 

If  28  is  20  per  cent  of  some  number  less  than  the 
number,  it  is  100  per  cent,  of  the  number — 20  per 
cent  of  the  number,  or,  it  is  80  per  cent,  of  the 
number. 

Insist  that  the  pupils  shall  not  use  the  term  per 
cent,  without  telling  "per  cent,  of  what." 


CHAPTER  XVI. 


LOSS   AND   GAIN. 

The  problems  in  percentage  that  give  most  trouble 
are  those  in  Loss  and  Gain ;  but  if  the  pupil  can 
always  answer  the  question  "per  cent,  of  what?"  not 
much  difficulty  should  be  encountered.  Every  pro- 
blem falls  under  one  of  the  general  problems  given. 

Permit  no  such  expressions  as  "let  100  per  cent.  = 
the  number." 

Such  solutions  are  algebraic  and  should  not  be 
tolerated  in  arithmetic. 

In  reading  the  problem  the  pupil  should  be  requir- 
ed frequently  to  supplement  the  test  by  expressing 
per  cent,  of  what.  Require  both  forms  of  solution 
C'34) 


Methods  in  Written  Arithmetic.  135 

but  use  the  one  that  is  most  convenient  for  the  par- 
ticular problem  under  consideration. 

The  most  troublesome  problems  are  of  the  second 
kind,  consequently  more  of  these  should  be  wrought 
than  of  the  others. 

ANALYSIS. 

1.  For  what  must  I  sell  a  horse  that  cost  me  $150 
to  gain  35  per  cent.? 

This  is  a  problem  of  the  first  kind.  I  must  sell 
the  horse  for  $150+35  Per  cent-  of  SI5o.  One  per 
cent,  of  $150  is,'etc. 

Or,  35  per  cent.  =  7-20.  I  must  sell  the  horse  for 
$150+7-20  of  $150,  etc. 

2.  Bought  sugar  at  8  cents  a  pound ;  the  wastage 
is  12^  per  cent.;  how  must  I  sell  it  so  as  to  gain  25 
per  cent.? 

This  is  a  problem  of  the  first  kind.  The  wastage 
is  12^  per  cent.,  or  ^,  of  each  pound  purchased; 
at  how  much  per  pound  must  I  sell  it  to  gain  25  per 
cent.,  or  ^,  of  the  cost  of  each  pound? 

Since  %  of  each  pound  is  lost,  only  ]fa  of  each 
pound  is  left,  hence  ^  of  a  pound  costs  8  cents.  ft 
of  a  pound,  then,  costs  1-7  of  8  cents,  or  8-7  cents, 


136  Methods  in  Writ  I  en  Arithmetic. 

and  8-8  of  a  pound  costs  64-7  cents,  or  9  1-7  cents. 
It  must  be  sold  for  9  1-7  cents  +  ^  of  9  1-7. 
etc. 

3.  I  buy  at  $7  and  sell  at  $8.50 ;  what  is  the  per 
cent,  of  gain? 

This  is  a  problem  of  the  second  kind.  If  I  buy  at 
$7  and  sell  at  $8.50  I  gain  $1.50.  $1.50  is  what  per 
cent,  of  $7  ?  Solve  by  both  methods. 

4.  I  buy   a   number  of  pounds   of  sugar.     The 
wastage  is  20  per  cent.,  and  it  is  sold  at  40  per  cent, 
above  cost ;  what  is  the  gain  per  cent.? 

This  is  a  problem  of  the  second  kind.  If  20  per 
cent.,  or  1-5,  of  the  sugar  is  lost,  the  remaining  4-5 
cost  as  much  as  the  whole ;  then  each  pound  remain- 
ing cost  5-4  of  the  original  price  per  pound.  If  each 
remaining  pound  is  sold  for  140  per  cent.,  or  7-5,  of 
the  original  price  per  pound  the  gain  on  each  pound 
is  7-5 — 5-4,  or  3-20,  of  the  original  price.  3-20  of 
the  original  price  is  what  per  cent,  of  5-4  of  the  orig- 
inal price  ? 

5.  I  bought  goods  for  $7.29.     At  what  price  must 
must  I  mark  them  that  I  may  fall   10  per  cent.,  lose 
10   per   cent,  in   bad   debts,  and   yet  gain   10  per 
cent.? 


Methods  in  Written  Arithmetic.  137 

Problems  of  this  kind  are  not  unusual.  The  only 
difficulty  is  "per  cent,  of  what  ?"  A  full  statement 
is  as  follows: 

I  bought  goods  at  $7.29;  at  what  price  must  I 
mark  them  that  I  may  gain  10  per  cent,  of  the  cost 
price,  lose  10  per  cent,  of  the  selling  price,  and  fall 
10  per  cent,  of  the  marked  price  ? 

If  the  gain  is  10  per  cent,  of  the  cost,  it  is  $.729 ; 
hence  $7.29-)-$  729-^8.019  is  collected.  Which  kind 
of  problem  is  this  part  ? 

If  10  per  cent.,  or  i-io,  of  the  selling  price  is  not 
collected,  then  $8.019  is  9  10  of  the  selling  price. 
This  part  is  a  problem  of  the  third  kind. 

If  $8.019  is  9-10  of  the  selling  price,  i-io  of  the 
selling  price  is  1-9  of  $8.019,  ar  $.891 ;  10-10  of  the 
selling  price  is  10  times  $.891,  or  $8.91. 

If  I  fall  i-io  of  the  marked  price,  $.91  must  be  9- 
10  of  the  marked  price,  etc. 

The  straight-line  analysis  should  be  employed 
where  it  is  possible  to  use  it.  Go  through  the  fore- 
going analysis  indicating  the  operation.  When  it  is 
completed  the  problem  will  have  assumed  the  follow- 
ing form: 


138  Methods  in   Written  Arithmetic. 

$7.29X11X10X10 


10X9X9 

Employing  cancellation   the   work,  r    materially 
shortened. 


CHAPTER  XVII. 


COMMISSION. 

Each  of  the  applications  of  percentage  furnishes 
problems  of  the  three  kinds. 

Those  in  commission  and  stock  investments  seem 
to  give  most  trouble. 

Commission  problems  may  be  characterized  as 
"  selling  problems"  and  "buying  problems,"  the  sec- 
ond being  the  more  difficult. 

The  old  question  "  Per  cent,  of  what?"  again  pre- 
sents itself  for  an  answer.  This  understood  the  diffi- 
culties vanish. 

A  commission  merchant  sells  for  a  customer  and 
is  to  receive  two  per  cent.  The  meaning  simply  is 
that  he  is  to  retain  two  per  cent,  of  all  moneys  that 
he  receives,  as  compensation  for  his  service*.  Few 


140  Methods  in  Written  Arithmetic. 

pupils  have  trouble  with  this  problem.  The  solution 
is  entirely  simple  when  the  problem  is  recognized. 
The  "  buying  problems,"  however,  perplex  the  dull 
pupils,  and  are  an  occasional  snare  to  the  bright 
ones. 

Money  is  sent  to  an  agent  to  be  invested  with  the 
understanding  that  the  amount  sent  covers  the  in- 
vestment and  the  agent's  commission.  The  books 
are  usually  silent  upon  the  latter  point,  leaving  it  to 
be  inferred  by  the  pupil. 

A  man  sends  a  commission  merchant  a  sum  of 
money  which  he  is  to  invest  at  3  per  cent.  What 
is  the  agent's  commission  and  how  much  did  he  in- 
vest ? 

The  statement  is  a  little  obscure,  because  the  lan- 
guage used  is  technical. 

There  are  two  methods  of  solution. 

(a)  For  each  dollar  which  the  agent  invests  he 
charges  3  cents  commission ;  hence  each  dollar  in- 
vested in  the  business  venture  uses  up  one  dollar 
and  three  cents  of  the  money  sent.  As  many  dollars 
will  be  invested  as  the  money  sent  is  times  one  dol- 
lar and  three  cents,  etc.  The  commission  is,  of 
course,  the  difference  between  the  money  sent  and 
the  money  invested. 


Mfthods  in   Written  Arithmetic.  141 

(<5)  The  commission  is  three  per  cent,  ol  the  mo- 
ney invested.  The  money  sent,  then,  is  103  per 
cent,  of  thenmountto  be  invested.  One  per  cent  of 
the  amount  to  be  invested  is  one  one-hundred-third 
of  the  amount  sent,  and  the  investment  is  100 — 103 
of  the  same  amount.  The  commission  is  3 — 103  of 
the  amount  sent,  or  3 — too  of  the  investment. 

By  this  method  the  commission  may  be  obtained 
without  finding  the  investment. 

An  exceedingly  useful  exercise  consists  in  the  class- 
ification of  problems.  Let  a  pupil  read  a  problem 
and  "  locate  it,"  without  stopping  to  solve  it.  The 
examples  in  several  different  arithmetics  may  be  ex- 
amined in  this  manner  and  readiness  acquired  by  the 
increased  familiarity. 

The  problems  in  commission  may  be  illustrated  in 
a  very  simple  manner.  Let  two  pupils  act  as  dealers 
and  athirdasacommission  merchant  or"middleman.'' 
Call  them  respectively  A,  B  and  C.  Pieces  of  paper 
suitably  prepared  may  represent  currency.  A  sends 
goods  to  C  to  be  sold  at  a  commission  of  two  per 
cent.  B  buys  them. 

From  every  dollar  that  C  receives  from  B,  he  takes 
two  cents  and  sends  the  remainder  to  A ;  hence  A 


142  Methods  in  Written  Arithmetic. 

receives  but  ninety  eight  per  cent,  of  what  his  goods 
sell  for  in  the  market. 

If  C  were  collecting  money  for  A  the  operation 
would  be  the  same.  This  form  of  the  problem  is  very 
simple,  and  is  easily  mastered. 

Suppose,  however,  that  A  sends  money  to  C  to 
invest  for  him  at  a  commission  of  two  per  cent.,  with 
the  understanding  that  the  money  sent  includes  the 
investment  and  the  commission  upon  it.  The  pupils 
are  inclined  to  think  that  the  commission  should  be 
reckoned  upon  the  whole  amount  sent,  and  subtracted 
from  it,  and  that  only  the  remainder  should  be  in- 
vested. 

They  do  not  remember  that  the  agent  is  to  get  two 
cents  for  investing  one  dollar,  not  for  investing  nine- 
ty-eight cents. 

Pupil  C  takes  one  dollar  from  the  money  that  A 
has  sent  him  and  buys  something  from  B.  To  pay 
himself  for  his  work  he  takes  two  cents  from  the  re- 
maining money  and  puts  it  into  his  pocket.  He  has 
now  invested  one  dollar  for  A,  but  has  used  up  $1.02 
of  his  money ;  so,  for  every  dollar  that  he  invests  he 
uses  up  $i-°2,  hence  he  can  invest  as  many  dollars 
as  the  money  sent  is  times  $1.02. 

For  the  other  form  of  solution  see  £,  p.  141. 


Methods  in   Written  Arithmetic.  143 

The  following  analysis  of  problems  found  in  Ray'a 
Higher  Arithmetic  will  illustrate  the  methods  : 

1.  Bought  flour  for  A;  my  whole  bill  was  $5802.- 
57,  including  charges  $76.85  and  commission  $148.. 
72;  find  the  rate  of  commission. 

The  whole  bill  must  include  the  investment,  the 
commission  upon  it,  and  the  remaining  charges. 
The  commission  and  charges  amount  to  $225.75.  The 
investment,  then,  is  $5802.57 — $225.57,  or  $5577 
The  question  now  is,  $148.72  is  what  per  cent,  of 
$5577  •'  Under  what  general  problem  does  this  fall? 
Work  by  either  method. 

2.  An  agent  sold  my  corn,  and,  after  reserving  his 
commission,  which  was  3  per  cent,  for  buying  and   3 
per  cent,  for  selling,  he  invested  the  remainder  in 
corn ;  his  charge  was  $12;  for  what  was  the  first  corn 
sold? 

Since  he  received  a  commission  of  3  per  cent,  of 
what  the  corn  sold  for,  only  97  per  cent,  of  this 
amount  remained.  Since  he  is  to  receive  3  per  cent, 
of  what  the  second  corn  costs,  97  per  cent,  of  what 
the  first  corn  brought  is  103  per  cent,  of  what  the 
second  cost.  One  per  cent,  of  what  the  second  cost 
is  1-103  of  97  per  cent,  of  what  the  first  brought; 
100  per  cent,  of  what  the  second  cost  is  100-103  of 


144  Methods  in  Written  Arithmetic. 

97  per  cent,  of  what  the  first  brought.  The  commis- 
sion on  the  second,  then,  is  3  103  of  97  per  cent,  of 
what  the  first  brought,  or  291-103  per  cent.,  or  2  85- 
103  per  cent,  of  what  the  first  brought.  The  commis- 
sion, then,  is  3  percent,  of  the  cash  of  the  first  plus 
2  85-103  per  cent,  of  the  same,  or  5  85-103  per  cent. 
$12,  then,  is  5  85-103  per  cent,  of  the  cost  of  the  first 
corn.  5  85-103  percent.  —  600-103  per  cent.  1-103 
per  cent,  of  what  the  first  cost  =  1-600  of  $12,  and 
100  per  cent,  of  the  cost  of  the  first  is  100x103  600 
ot  $12,  or,  omitting  common  factors, 


CHAPTER  XVIII. 


STOCK    INVESTMENTS. 

This  is  another  of  the  topics  that  trouble  pupils. 

The  same  set  of  problems  recurs.  All,  however, 
are  assignable  to  one  or  another  of  the  Three  General 
Problems.  The  trouble  here  as  elsewere  is  to  answer 
the  old  question  "Per  cent,  of  what?" 

The  meaning  of  all  terms  used  must  be  entirely 
comprehended. 

Since  there  are  five  terms  in  common  use  we  may 
expect  five  problems. 

These  terms  are  Investment,  Market  Value,  Rate 
of  Interest,  Rate  of  Income,  Income. 

Most  of  these  problems  are  quite  simple.  Those 
are  most  difficult  that  fall  under  the  Second  General 
Problem.  The  following  will  illustrate. 


146  Methods  in   Written  Arithmetic. 

vVhat  is  the  rate  of  incomp  pn  6  per  cent,  bonds 
bought  at  T.  12  ? 

These  bonds  pay  as  interest  6  per  cent,  of  their 
par  value.  They  arc  bought  at  112  per  cent,  of  their 
par  value.  The  question  is  6  per  cent,  of  a  numbe; 
is  what  per  cent,  of  112  per  cent,  of  the  same 
number? 

If  pupils  do  not  grasp  the  subject  take  a  special 
case. 

I  buy  a  hundred  dollar  six  per  cent,  bond  at  1 1 2 
per  cent,  of  its  par  value.  What  rate  of  interest  do  1 
receive  on  my  investment? 

The  bond  cost  $112.  It  pays  $6  a  year  as  interest 
$6  is  what  .per  cent,  of  $1  12  ? 

Problems  like  the  following  are  sometimes  trouble- 
some: 

At  what  rate  must  4  per  cent,  bonds  be  bought  to 
make  the  investment  pay  10  per  cent.? 

The  question  is,  4  per  cent,  of  anything  is  10  per 
cent,  of  what  per  cent,  of  it  ?  or  4  is  10  per  cent,  of 
what? 

If  4  per  cent,  of  the  par  value  is  10  per  cent,  of  the 
investment  i  per  cent,  of  the  investment  is  i-io  of  4 
per  cent,  of  the  par  value,  or  4-10  per  cent,  of  the  par 
value,  etc. 


Methods  in    Written  Arithmetic.  147 

This  may  also  be  made  simpler  by  making  a 
special  problem. 

At  what  per  cent,  of  the  par  value  must  I  buy  a 
4  per  cent.  $100  bond  that  I  may  receive  10  per 
cent,  interest  on  my  investment  ? 

This  bond  pays  $4  a  year  as  interest.  $4,  then,  is 
10  per  cent,  of  the  price  to  be  paid  for  it.  i  per 
cent,  of  the  price  to  be  paid  for  it  is  i-io  of  $4  or 
40  cents.  100  per  cent,  of  the  price  to  be  paid  for  it 
is  100  times  40  cents,  or  $40.  $40  dollars  is  40  per 
cent,  of  $  100,  hence  I  must  buy  the  bond  for  40  per 
cent,  of  the  par  value,  or  60  per  cent,  discount. 

The  following  problem  from  Ray's  Higher  Arith. 
illustrates  the  necessity  of  knowing  "per  cent,  of 
what.'* 

Suppose  10  per  cent.  State  Stock  20  per  cent, 
better  in  market  than  4  per  cent,  railroad  stock;  if 
A's  income  be  $500  from  each,  how  much  money 
has  he  paid  for  each,  the  whole  investment  bringing 
6  2-333  Per  cent.? 

If  his  annual  income  from  the  state  stock  is  $500, 
t'len  $500  must  be  10  percent,  of  the  par  value  of 
his  state  stock.  If  $500  is  i-io  of  the  par  value  then 
the  par  value  is  $5000. 


148  Methods  in  Written  Arithmetic. 

Reasoning  similarly  he  must  have  $12,500  of  the 
railroad  stock. 

If  the  market  value  of  the  state  stock  is  20  per 
cent,  of  the  market  value  of  the  railroad  stock  more 
than  the  market  value  of  the  railroad  stock  it  is  6-5 
of  it.  If  the  market  price  of  the  two  were  the  same 
the  state  stock  would  cost  2-5  as  much  as  the  rail- 
road stock,  since  these  are  2-5  as  much  of  it.  Bui 
since  it  cost  6-5  as  much  in  market,  his  state  stock 
must  have  cost  6-5  of  2-5,  or,  12-25  as  much  as  the 
railroad  stock. 

The  two,  then,  must  have  cost  25-25+12-25  01 
37-25  as  much  as  the  railroad  stock. 

His  whole  income  is  $1000.  This  is  6  2-333  Per 
cent,  of  his  investment.  6  2-333  per  cent.  =  2000- 
333  Per  cent.  If  $1000  is  2000-333.  per  cent  of  his 
investment  1-333  Per  cent,  of  his  investment  is 
i  2000  of  |iooo,  or  $%.  333-333  percent  or  one  per 
cent,  of  the  investment  is  333  times  $}4,  or  $166.50, 
andioopercent.oftheinvestmentis  100  times  $166.50 
or  $16650  But  the  whole  investment  is  37-25  of  the 
cost  of  the  railroad  stock,  hence  1-25  of  the  cost  of 
the  railroad  stock  is  1-37  of  $16650=^450  and  25-25= 
25  times  $450,  or  $11250. 

What  is  the  cost  of  the  other  ? 


Methods  in   Written  Arithmetic.  149 

is  may  be  shortened  by  indicating  the  operations 
as  the  analysis  progresses,  and  omitting  common  fac- 
tors at  the  close. 

$1000X333X100x25 

a  000x3? 


CHAPTER  XIX. 


INTEREST. 

The  computation  of  interest  affords  one  of  the 
most  common  applications  of  the  problems  of  per- 
centage. The  pupils  should  understand  thoroughly 
what  is  meant  by  "ten  per  cent,  per  annum."  It 
means  simply  that  for  the  use  of  a  given  thing,  what- 
ever it  may  be,  for  one  year,  ten  per  cent,  of  that 
thing  is  to  be  allowed.  For  fractional  parts  of  a  year, 
then,  fractional  parts  of  ten  per  cent,  of  the  given 
thing  are  to  be  paid,  etc.;  nence  these  problems  are 
of  the  first  kind. 

Our  text-books  present  several  methods  of  com- 
puting interest.  One,  made  perfectly  familiar,  is 
sufficient,  and  is  immeasurably  better  than  a  super- 


Methods  in   Written  Arithmetic.  151 

ficial  knowledge  of  three  or  four.  "A  ten  per  cent, 
method  "  has  found  its  way  into  few  or  none  of  the 
books.  Those  unfamiliar  with  it  will  judge  of  its  worth 
after  examining  the  following  statements : 

1.  The  interest  for  one  year  is  found  by  moving 
the  decimal  point  one  place  to  the  left. 

2.  Since  three  hundred  and  sixty  days  are  con- 
sidered a  year,  in  computing  interest  ordinarily,  it  fol- 
lows that  the  interest  for  thirty  six  days  is  one  hun- 
dredth of  the  principal,  and  is  found  by  moving  the 
decimal  point  two  places  to  the  left. 

3.  The  pupil,  then,  must  always  keep  in  mind  these 
two  periods — one  year,  and  thirty-six  days.    For  any 
number  of  years  multiply  one-tenth  of  the  principal 
by  the  given  number ;  for   any  number  of  months 
take  parts  of   the  interest  for  one  year;    for  any 
number  of  days  take  parts  of  the  interest  for  thirty- 
six  days. 

For  rapid  and  accurate  work  "a  method"  is  neces- 
sary and  should  be  adhered  to  rigidly.  The  following 
illustrative  problems  will  indicate  the  methods  in  all 
cases: 


152  Methods  in  Written  Arithmetic. 

$2.6.5.80  i  year.  8  mos.  28  days. 
2  6.5  80 
8.8  6 

8.8  6  4  12 
.8  86  4  12 
.8  86  4 

•2  95 


$46.37 

The  time  is  more  than  one  year  and  less  than  two. 
In  such  cases : 

First.  Move  the  decimal  point  one  place  to  the 
left. 

Second.  Draw  a  horizontal  line  below  the  number 
of  months,  and  separate  this  number  into  convenient 
divisors  of  twelve. 

Third.  Take  such  parts  of  the  interest  for  one  year 
as  the  divisors  are  of  twelve,  and  write  the  results  for 
addition  to  the  interest  for  one  year. 

Fourth.  Draw  a  horizontal  line  below  the  number 
of  days  and  separate  the  number  into  convenient 
divisors  of  thirty-six. 

Fifth.  Indicate  one-tenth  of  tne  interest  for  one 
year  and  take  such  parts  of  this  result  as  the  divisors 
are  of  thirty-six. 


Methods  in  Writtcri  Arithmetic.  153 

Use  no  unnecessary  figures.  Observe  the  model 
carefully. 

2.  If  the  time  is  less  than  one  year : 

First.  Find  one-tenth  of  the  principal  and  draw  a 
horizontal  line  below  it. 

Second.  Separate  months  and  days  as  before,  ob- 
tain results  in  same  way  and  write  them  below  the 
line  for  addition. 

$4.8.6.50        10  months.         14  days. 


2  4.3  25         6  12 

i  6.2  16         4  a 

1.6  21 

.2  70 


3.    If  the  time  is  two  years  or  more: 

First.  Find  one-tenth  of  principal  and  draw  a  hori- 
zontal line  below  it 

Second.  Below  this  line  write  as  many  times  the 
interest  for  one  year  as  there  are  years. 

Third.  For  months  and  days  proceed  as  before. 


154  Methods  in  Written  Arithmetic. 

$5.2.6.70     4  years.     7  months.     17  days. 


21  0.6  8  6  12 

2  6.3  35  i  4 

4-3  89  i 

'•7  55 
•5  85 
.1  46 


$243.89 

Observe  that  in  the  above  final  results  no  mills 
are  written.  The  sum  of  the  mills  is  found  and  in 
reducing  it  to  cents,  a  remainder  of  five  or  more  is 
called  one  cent.  A  smaller  remainder  is  disre- 
garded. 

If,  in  the  final  result,  the  decimal  point  be  re- 
moved one  place  to  the  left,  the  interest  for  the 
given  time  at  one  per  cent,  is  obtained.  This 
multiplied  by  the  given  rate  will  give  the  result 
required. 

The  only  objection  to  the  ten  per  cent,  method  as 
a  general  method  is  that  in  problems  involving 
several  calculations,  as  in  partial  payments,  pupils 
forget,  sometimes,  to  find  the  interest  at  the  required 
rate  after  finding  it  at  10  per  cenx. 


Methods  in   Written  Arithmetic.  155 

A   GENERAL  PROBLEM. 

Find  the  interest  on  the  principal  for  one  year  at 
the  given  rate.  If  the  rate  is  8  per  cent.,  take  8  hun- 
dredths  of  the  principal ;  a  similar  process  is  em- 
ployed for  any  other  rate.  In  all  other  respects  the  op- 
erations are  the  same  as  in  the  ten  per  cent,  method. 
To  find  the  interest  for  any  number  of  years  multiply 
the  interest  for  one  year  by  the  given  number  of 
years.  Separate  the  months  into  convenient  divisors 
of  twelve  and  take  such  a  part  of  the  interest  for  a 
year  as  the  number  of  months  is  of  twelve.  Take 
one  tenth  of  the  interest  for  a  year  and  the  interest 
for  thirty-six  days  is  obtained.  Separate  the  number 
of  days  into  aliquot  parts  of  36  and  proceed  as 
before. 

SIX-PER-CENT.    METHOD. 

Since  any  principal  at  6  per  cent,  gains  6  hun- 
dredths  of  itself  in  1 2  months  it  will  gain  one  hun- 
dredth of  itself  in  2  months  and  will  double  itself  in 
200  months.  It  will,  therefore,  gain  one  thousandths 
of  itself  in  6  days,  or  one  tenth  of  two  months.  The 
period  of  time  to  be  kept  in  mind  with  especial  care 
are  200  months,  2  months  and  6  days,  the  principal 


156  Methods  in  Written  Arithmetic. 

doubling  itself,  gaining  one  hundredth     and    one 
thousandth   of  itself  in  the  respective  periods. 

FORM   OF  WORK. 

Reduce  the  years  to  months  and  to  this  result  add 
the  given  months.  Draw  a  line  below  the  numbei 
and  separate  it  into  divisors  of  200.  Separate  the 
days,  in  the  same  manner,  into  divisors  of  60  days. 

Problem.  What  is  the  interest  on  $5680  for  5  yrs. 
5  months  and  22  days? 

$5680    5  years,  5  months,  =  65  months.  22  d. 


1420.  50  20 

284.  xo  2 

142.  5 

iS-933 
1.893 


$1866.83 

An  explanation  similar  to  the  following  will  be 
found  useful : 

5  yrs.  and  5  mos.  equal  65  mos.,  which  I  separate 
into  50  mos.,  10  mos.  and  5  mos.  In  50  mos.  the 
principal  will  gain  one-fourth  of  itself;  in  10  mos.  it 


Methods  in  Written  Arithmetic.  157 

will  gain  one-fifth  cf  this  amount;  in  5  mos.  it  will 
gain  one-half  of  the  interest  for  10  mos.  I  separate 
22  days  into  20  days  and  two  days.  Since  20  days  is 
y$  of  60  days,  the  interest  is  J/z  of  i-ioo  of  the  prin- 
cipal ;  the  interest  for  2  days  is  i-io  of  the  interest 
for  20  days.  Combining  these  several  results,  etc. 

Facility  cannot  be  expected  until  questions  like 
the  following  can  be  answered  with  perfect  readiness. 

The  principal  gains  what  part  of  itself  in  ico  mos.? 
20  mos.?  66^4  mos.?  33^/3  mos.?  20 days?  3  days? 

TO    FIND   THE   TIME   BETWEEN    TWO   DATES. 

Find  the  full  number  of  years,  the  full  number  of 
months,  and  count  the  remaining  days. 

The  following  illustrations  will  show  the  forma 
which  the  problem  may  assume : 

1.  What  is  the  time  between  Jan.  5, 1874,  and  April 
56,  1878? 

METHOD. 

From  Jan.  5,  '74,  to  Jan.  5,  '78,  is  4  years. 
From  Jan.  5,  '78,10  Apr.  5/78,  is  3  months. 
From  Apr.  5,  '78,  to  Apr.  16,  '78,  is  n  days. 

2.  What  is  the  time  between  March  15,  '72,  and 

Feb.  20,  '76 

METHOD. 

From  March  15,  '72,  to  March  15,  '75,  is  3  years. 


158  Methods  in  Written  Arithmetic, 

From  March  15,  '75,  to  Feb.  15,  '76,  is  n  mos. 
From  Feb.  15,  '76,  to  Feb.  20,  '76,  is  5  days. 
3   What  is  the  time  between  Sept.  20/68,  and  Aug. 
10, '73? 

METHOD. 

From  Sept.  20,  '68,  to  Sept.  20,  '72,  is  4  years. 
From  Sept.  20,  '72,  to  July  20,  '7 3,  is  10  months. 
From  July  20,  '73,  to  Aug.  10,  '73,  is  21  days. 

PARTIAL    PAYMENTS. 

Pupils  should  be  so  instructed  that  they  shall  be 
able  to  prepare  notes  of  any  form  promptly,  neatly 
and  accurately.  Punctuation,  legibility,  form  of  pa- 
per, and  all  the  details  should  receive  special  atten- 
tion. Notes  should  be  written  with  ink  and  handed 
to  the  teacher  for  criticism. 

The  reasonableness  of  the  U.  S.  Rule  should  be 
apparent. 

Much  practice  in  performing  problems  in  partial 
payments  is  needed  before  facility  and  accuracy  can 
be  acquired. 

Some  definite  form  should  be  selected  and  insisted 
upon.  The  following  is  suggested  : 

$725.50.  Bloomington,  III. ,  Feb.  15,  1880. 

Two  years  after  date,  for  value  received,  I  promise 
to  pay  Edwin  C.  Hewett,  or  order,  Seven  Hundred 


Methods  in  Written  Arithmetic.  159 

Twenty-Five  and   50-100  Dollars,  with  interest  at 
eight  per  cent,  per  annum  from  date. 

JOHN    SMITH. 
Indorsements  : 

July  20,  1880,  $50.00 
Sept.i2,  1880,    62.50 
Jan.     6,  1880,    42.00 
Dec.  24,  1881,      6.00 
Feb.    i,  1882,  100.00 
What  was  due  Feb.  15,  i8Cc  ? 
With  this  problem  before  the  pupils,  have  them 
rut  it  upon  the  board,  slate  or  paper,  in  the  following 
form  : 
$725-5° 


$  50.00. 
$  62.50 
$  42.00. 
$     6.00. 
$100.00. 


In  solving  the  problem  use  the  form  given  in  the 
ordinary  tjxt  books.  No  multiplication  should  be 
permitted  except  in  obtaining  the  interest  for  a  year. 
Reject  all  results  below  tenths  of  mills. 


8  per  cent. 

Feb.  15, 

I  ooO 

u 

d. 

July  20, 

iSSD 

1  5' 

5 

Sept.  12, 

1880 

i  l: 

23 

Jan.  6, 

1881 

\  3" 

25 

Dec.  24, 

1881 

i  ir- 

s 

18 

Feb.  i, 

1882 

i  '• 

Feb.  15, 

1882 

i 

CHAPTER  XX. 


PRESENT  WORTH. 

The  present  worth  of  a  sum  of  money  due  at  a 
future  time,  and  not  bearing  interest^  is  that  sum 
which  put  at  interest  at  the  given  rate  will  amount 
to  the  obligation  in  the  given  time. 

There  are  two  methods  of  solution. 

i.  What  is  the  present  worth  of  a  debt  of  two 
hundred  dollars,  due  in  two  years  and  six  months, 
and  not  bearing  interest,  money  being  worth  8  per 
cent,  per  annum  ? 

(a]  If  one  dollar  be  put  at  interest  at  8  per  cent, 
for  two  years  and  6  months  it  will  amount  to  $1.20. 
One  dollar,  then,  is  the  present  worth  of  $  1.20,  due 
in  two  years  and  six  months.  The  present  worth  of 
$200,  then,  due  at  the  same  time  must  be  as  many 
(160) 


Methods  in   Written  Arithmetic.  161 

dollars  as  there  are  times  $1.20  in  $200.  There  are 
166.66  times  $1.20  in  $200;  hence  the  pressent 
worth  of  $200  under  the  above  conditions  is  $166.66. 

The  justice  of  the  above  is  obvious.  If  I  owe  a 
debt  which  is  not  due  for  two  and  a  half  years  and  I 
am  not  to  pay  interest  upon  it,  I  am  entitled  to  the 
use  of  the  amount  for  that  time.  If  I  pay  the  debt 
in  full  before  its  maturity  I  lose  the  use  of  the  mo- 
ney for  the  given  time.  If,  however,  I  pay  the  cre- 
ditor a  sum  which  he  can  put  at  interest  at  the  given 
rate  for  the  given  time  and  which  will  amount  to  the 
debt  at  its  maturity  neither  loses. 

The  difference  between  the  present  worth  and  the 
amount  of  the  debt  is  the  Discount. 

(<£)  A  second  method  of  solution. 

Any  sum  of  money  put  at  interest  at  8  per  cent, 
for  two  and  a  half  years  will  amount  to  six-fifths  of 
itself.  The  amount,  then,  is  six-fifths  of  its  present 
worth.  One-fifth  of  the  present  worth  is  one-sixth 
of  the  amount.  Five  fifths  of  the  present  worth 
equals  five-sixths  of  the  amount:  hence,  in  this  case, 
the  present  worth  is  five-sixths  of  $200. 

As  will  be  seen,  the  example  falls  under  the  third 
General  Problem  of  Percentage. 

This  method  enables  pupils  to  obtain  the  discount 


1 62  Methods  in  Written  Arithmetic. 

without  finding  the  present  worth.  The  [.•Nis.ent 
worth  is  five-sixths  of  the  amount.  The  discount* 
then,  is  one-sixth  of  the  amount. 

Suppose  the  debt  is  bearing  interest.  If  so,  find 
the  amount  which  is  to  be  paid  at  the  maturity  of  the 
note  and  proceed  as  above. 

True  Discount  is  one  of  the  difficult  applications 
of  percentage. 


CHAPTER  XXI. 


BANK   DISCOUNT. 

The  chief  cause  of  difficulty  in  this  and  kindred 
topics  is  th,j  ignorance  of  pupils  in  regard  to  the 
practical  details  of  business  matters.  Although  nine- 
ty-five per  cent,  of  Ae  business  of  the  country  is  car- 
ried on  by  means  of  checks  and  drafts,  yet  few  pupils 
have  seen  either.  Teachers  should  instruct  the  pu- 
pHs  in  such  matters  m  order  that  the  work  of  the 
class-room  may  have  that  air  of  reality  without  which 
little  good  is  reached. 

Many   teachers   are   as  ignorant  as  their  pupils. 

They  must  inform  xhemselves  if  they  expect  to 
succeed  in  making  these  ^natters  clear.  Obtain  ac- 
curate and  clear  answers  to  the  following  ques- 
tions : 


164  Methods  in   Written  Arithmetic. 

1.  What  is  a  bank  ? 

2.  How  many  kinds  of  banks  are  there  ? 

3.  By  what  other  name  are  banks  of  issue  desig- 
nated? 

4.  Suppose   a  bank   of   issue  should  fail,  would 
the  holders  of  its  notes  lose  anything?  Why? 

5.  Of  what  advantage  is  a  bank  of  deposit? 

6.  In  how  many  ways  may  money  be  deposited  in 
a,  bank? 

7.  What  is  a  certificate  of  deposit  ?  Write  one. 

8.  If  one  has  deposited  money  and   has  taken  a 
certificate,  how  can  he  get  the  money? 

9.  Can  anyone  else  get  it  ?  If  so,  how  ? 

10.  What  is  a  "pass  book?"    What  use  is  made 
of  it. 

11.  What  is  a  check?  Write  one. 

Banks  can,  of  course,  derive  no  profit  from  deposits 
unless  they  can  use  them.  Experience  demonstrates 
that  they  may  safely  lend  a  certain  part  of  their  de- 
posits and  yet  be  able  to  meet  all  ordinary  demands 
upon  them;  hence  lending  money  is  the  chief  busi- 
ness of  a  bank.  Bankers  conduct  this  business 
somewhat  differently  from  private  individuals.  The 
borrower  gives  a  note  for  a  certain  lime,  without  in- 
terest until  due.  The  banker  calculates  the  interest 


Methods  in  Written  Arithmetic.  165 

upon  the  face  of  the  note  for  the  given  time  and  for 
three  days  additional,  called  "  days  of  grace,"  sub- 
tracts this  interest  from  the  face,  and  gives  the  bor- 
rower the  remainder. 

This  is  called  discounting  notes. 

BANK  DISCOUNT  presents  two  problems,  the  sec- 
ond of  which  is  the  more  difficult.  There  is  no  rea- 
son, except  custom,  why  the  first  should  differ  from 
the  True  Discount.  It  results  in  an  advantage  to  the 
banker,  and  is  a  trifle  easier  to  compute. 

(a)  To  find  the  present  worth  of  a  note  discounted 
at  a  bank,  and  not  bearing  interest,  find  the  simple 
interest  on  the  face  of  the  note  for  the  given  time  plus 
three  days,  and  subtract  it  from  it.  The  banker  thus 
receives  a  higher  rate  than  the  one  specified.  For 
example : 

Suppose  the  face  of  the  note  to  be  $200,  the  time 
60  days,  and  the  rate  6  per  cent.  The  interest  on 
$200  for  63  days  is  $2.10.  The  avails,  then,  are 
$197.90.  The  banker  pays  that  amount  for  the  note. 
At  its  maturity  the  debtor  gives  $200  for  the  same 
note,  thus  paying  $2.10  interest  on  $197.90  for  63 
days ;  the  rate,  then,  is  something  more  than  6  pel 
cent. 

It  is  a  common  saying  that  the  interest  in  such 


1 66  Methods  tn  Written  Arithmetic. 

cases  is  paid  in  advance.  This  is  obviously  incor- 
rect since  the  debtor  pays  nothing  until  the  maturity 
of  the  note.  He  simply  pays  a  rate  of  interest  a 
little  higher  than  the  nominal  rate. 

The  second  problem,  as  has  been  stated,  is  more 
confusing. 

(£)  To  find  the  face  of  a  note  when  the  avails  are 
given. 

For  what  sum  must  a  note  be  given  that  will 
yield  $260  when  discounted  at  6  per  cent,  for  90 
days? 

The  simple  interest  on  one  dollar  for  93  days  at  6 
per  cent,  is  $.0155.  A  note  whose  face  is  one  dollar 
consequently,  will  yield  $.9845.  To  yield  $260  the 
face  of  the  note  must  be  as  many  times  one  dollar, 
as  $260  is  times  $.9845,  etc. 

This  problem  may  also  be  performed  by  the  second 
method  employed  in  True  Discount. 

Problems  like  the  following  are  very  helpful  in 
teaching  pupils  the  philosophy  of  this  kind  of  work 
The  old  question  "Per  cent,  of  what  ?"  is  the  one 
that  is  constantly  presented. 

i.  What  is  the  rate  of  interest  per  -annum  on  a 
30  day  note  when  discounted  at  i  per  cent,  a  month? 

The  discount  is  .on  of  the  face  of  the  note.     The 


Methods  in  Written  Arithmetic.  167 

avails  are  .989  of  the  face  of  the  note.  The  interest 
is  11-989  of  the  avails.  The  time,  however,  is  33 
days.  For  3  days  it  would  be  i-n  of  11-989  of  the 
avails;  for  30  days  it  would  be  10 times  this  amount, 
and  for  360  days,  12  times  this  amount. 
The  work  may  be  shown  in  the  following  form : 

11X10X12  120 

The  resulting   fraction   is  which 

989X11  989 

changed  to  per  cent,  equals — what  ? 

This,  like  some  of  the  problems  previously  noticed, 
may  be  made  clearer  by  a  special  case. 

Suppose  the  note  were  for  $100.  The  discount, 
then,  would  be  $1.10.  The  avails  would  be  $98.90- 
The  person  selling  the  note  would  receive  $98.90  for 
it.  When  the  note  is  paid,  the  person  discounting 
it  would  receive  $100  for  it;  that  is,  $1.10  more  than 
he  paid  for  it.  This  $1.10,  consequently,  is  the  in- 
terest on  $98.90  for  33  days.  At  the  same  rate  the 
interest  for  three  days  would  be  i-n  of  $1.10,  orl.io 
For  360  days  it  would  be  120  times$.ioor  $12. 

If  $12  be  paid  for  the  use  of  $98.90  for  one  year 
what  is  the  rate  ? 

(2)  What  rate  of  discount  on  a  60  day  note  will 
yield  6  per  cent,  per  annum  ? 


1 68  Methods  in  Written  Arithmetic. 

Since  the  interest  is  6  per  cent,  of  the  avails  for 
360  days,  it  is  twenty-one  twentieths  of  one  per 
cent,  of  the  avails  for  63  days.  The  face  of  the 
note  is  101.05  Per  cent,  of  the  avails  and  the  avails 
10,000-10,105  of  the  face  of  the  note.  The  discount 
is  105-10,105  of  the  face  of  the  note.  This  being 
the  discount  for  63  days  it  is  120-21  of  this  for  360 
days. 

Reduce  and  change  to  per  cent. 

SPECIAL    PROBLEM. 

Suppose  the  above  note  were  for  $ioo.  What 
must  be  the  rate  of  discount  to  make  it  yield  interest 
at  the  rate  of  6  per  cent,  on  the  avails  ? 

$100  is  the  sum  of  the  avails  of  the  note  and  6  per 
cent,  interest  on  the  avails  for  63  days. 

Since  any  principal  at  6  per  cent,  amounts,  in  63 
days,  to  101.05  per  cent.,  or  10105-10000  of 
itself,  $100  must  be  10,105-10,000  of  the  avails, 
i  10000  of  the  avails  is  1-10105  °f  $i°o>  and 
10000-10000  of  the  avails  must  be  10000-10105 
of  $100.  The  discount  then  is  105-10105  of  $100. 
Since  this  is  the  discount  for  63  days,  for  3  days  it  is 
i-2i  of  105-10105  of  a  $100,  or  canceling  1-201,  of 
$100.  For  360  days  it  is  120  2021  of  $100.  120-2021 
=  what  per  cent.? 


Methods  in   Written  Arithmetic.  169 

Another  kind  of  business  transacted  by  banks  is 
the  collection  of  debts. 

A  in  one  place  has  a  bill  against  B  in  another.  A 
gives  his  bank  an  order  upon  B.  The  bank  sends 
this  order  to  a  bank  in  the  town  in  which  B  resides 
and  it  is  presented  for  payment.  Since  B  is  anxious 
to  maintain  a  reputation  for  prompt  payment  he  is 
more  likely  to  pay  the  bill  than  if  it  were  sent  directly 
to  him  by  A.  In  such  a  case  A  is  said  to  "draw 
upon"  B.  If  B  should  refuse  to  pay  the  draft  the 
collector  would  go  to  a  notary  and  make  an  affidavit 
to  that  effect.  The  affidavit  would  be  attached  to 
the  note  and  returned  to  the  bank  sending  it.  The 
draft  is  then  said  to  be  "protested."  Business  men 
are  naturally  reluctant  to  have  their  bills  go  to 
"protest"  unless  there  is  a  good  reason  for  it. 

A  third  kind  of  business  in  which  banks  engage  is 
the  sale  and  purchase  of  Bills  of  Exchange. 


CHAPTER  XXII. 


EXCHANGE. 

Suppose  that  A  in  Bloomington,  desires  to  pay  B, 
in  New  York,  a  sum  of  money;  he  may  do  this  in 
any  one  of  several  ways. 

1.  He  may  send  the  money  by  express. 

2.  He  may  send  postal  money  orders. 

3.  He  may  buy  a  bill  of  exchange. 

Let  us  see  how  a  system  of  exchange  may  grow  up 
between  two  cities. 

Suppose  that  A  can  find  some  one  in  Bloomington 
to  whom  some  one  in  New  York  owes  an  amount 
equal  to  the  amount  that  A  owes  B.  A  can  purchase 
of  him  an  order  on  this  New  York  creditor  and  send 
it  to  B.  B  can  receive  his  pay  on  presentation  of 
this  order.  Both  debts  are  thus  discharged  without 
the  double  transmission  of  money.  The  convenience 
of  this  method  is  apparent. 


Methods  in   Written  Arithmetic.  171 

Since  it  is  not  possible  always  to  find  persons  in 
the  relations  supposed  above,  a  banker  in  Blooming- 
ton  and  another  in  New  York  arrange  to  pay  orders 
that  each  may  draw  upon  the  other. 

All  that  is  necessary  now  is  for  A  to  purchase  of 
his  bank  an  order  upon  its  New  York  "Correspond- 
ent" and  sent  it  to  B,  who  presents  it  and  receives 
his  pay. 

The  New  York  bank  in  turn  may  sell  orders  upon 
tne  Bloomington  bank. 

This  is  the  essence  of  all  systems  of  exchange, 
whether  domestic  or  foreign. 

These  orders  are  called  Drafts,  or  Bills  of  Ex- 
change. 

If  the  Bloomington  bank  sells  more  drafts  upon 
the  New  York  bank  than  it  pays  for  it,  money  must 
be  sent  to  balance  the  account. 

What  is  the  difference  between  a  check  and  a 
draft  ?  What  is  the  difference  between  a  domestic 
draft  and  a  foreign  draft  ? 

Drafts  are  of  two  kinds — Sight  and  Time.  A  Sight 
Draft  is  payable  upon  presentation. 

A  Time  Draft  is  payable  a  specified  time  after 
"sight,"  that  is,  after  the  draft  has  been  presented 
and  the  proper  officer  has  written  across  the  face  the 


172  Methods  in  Written  Arithmetic. 

word  accepted,  the  date  of  acceptance,  and  the  signa- 
ture of  the  banking  company. 

A  time  draft  should  be  cheaper  than  a  sight  draft 
for  the  same  amount,  since  the  seller  has  the  use  of 
the  purchaser's  money  for  a  specified  time  before  the 
draft  is  redeemed. 

When  will  sight  drafts  be  at  a  premium?  When  at 
a  discount? 

Exchange  presents  the  same  problems  as  Bank 
Discount. 

1.  What  is  the  cost  of  $3600   sight  exchange  on 
New  York  at  i-io  per  sent,  premium? 

The  draft  will  cost  $3600  plus  i-io  per  cent,  of 
$3600. 

i  per  cent,  of  $3600  is  $36.  i-io  per  cent,  of  §3600 
is  i-io  of  $36,  or  $3.60.  The  draft  will  cost 
$3603.60. 

If  exchange  were  at  a  discount  instead  of  a  pre- 
mium the  method  would  be  the  same. 

2.  How  large  a  sight  draft  on  New  York  will  $3600 
buy,  exchange  being  i-io  per  cent,  premium? 

(a)  A  draft  for  one  dollar  will  cost  one  dollar  plus 
i-io  per  cent,  of  a  dollar,  or  $1.001.  $3600  will  buy 
a  draft  whose  face  is  as  many  times  one  dollar  aa 
$3600  is  times  $1.001. 


Methods  in   Written  Arithmetic.  173 

(£)  $3600  is  the  face  of  the  draft  plus  i-io  per 
cent,  of  the  face.  It  is,  then,  100.1  per  cent.,  or 
looi-iooo  of  the  face  of  the  draft. 

i-iooo  of  the  face  of  the  draft  is  i-iooi  of  $3600. 
i ooo- 1 ooo  of  the  face  of  the  draft  is  1000-1001  of 
$3600. 

The  same  problems  occur  with  time  drafts,with  the 
added  element  of  bank  discount. 

1.  What  will  a  6o-day  draft  for  $6000  cost  at  y± 
per  cent,  premium,  interest  at  6  per  cent.? 

(fl)  Since  this  draft  is  not  payable  for  63  days, 
the  purchaser  should  be  allowed  interest  on  his  money 
for  that  length  of  time.  It  is  customary  to  discount 
the  draft  for  the  time  it  has  to  run.  If  the  exchange 
were  neither  at  discount  nor  premium  the  purchaser 
would  pay  for  the  draft  $6000  minus  the  interest  on 
$6000  for  63  days  at  6  per  cent.,  or  $5937-  Since 
exchange  is  at  ^  per  cent,  premium,  he  must  pay 
$5937  plus  3/i  Per  cent,  of  $6000,  or  $5982. 

(^)  A  draft  for  one  dollar  will  cost  one  dollar  plus 
y±  per  cent,  of  a  dollar  minus  the  bank  discount  of  a 
dollar  for  63  days  at  6  per  cent.  A  draft  for  $6000 
will  cost  6000  times  this  amount. 

2.  What  is  the  face  of  a  90  day  draft  that  may  be 


1 74  Methods  in  Written  Arithmetic. 

purchased  for  $4320.50,  exchange  ^  per  cent,  dis- 
count, interest  6  per  cent.? 

(a)  A  draft  whose  face  is  one  dollar  will  cost  one 
dollar  minus  l/?,  per  cent,  of  a  dollar,  minus  the  bank 
discount  of  one  dollar  for  93  days ;  this  equals 
£-98325  ',  $4320.50  will  buy  a  draft  whose  face  is  as 
many  dollars  as  $4320.50  is  times  $.98325,  which  is, 
etc. 

(£)  The  bank  discount  of  any  principal  for  93  days 
at  6  per  cent,  is  .0155  of  that  principal.  The  ex- 
change discount  is  ^  per  cent.,  or  .00125  of  the 
principal.  Their  sum  is  .01675, — the  entire  discount. 
The  draft  will  cost  its  face  minus  .01675  of  its  face,  or 
.98325  of  its  face;  but  the  draft  is  to  cost  $4320.50. 
$4320.50,  then,  is  .98325  of  the  par  value  of  the  draft. 
.00001  of  the  par  value  is  1-98325  of  $4320.50,  and 
the  par  value  is  100,000  times  this  amount. 

(3)  A  30-day  draft  for  $4000  costs  $4010,  interest 
6  per  cent.:  what  is  the  rate  of  exchange? 

The  bank  discount  of  $4000  for  33  days  is  $22. 
The  premium  is  the  sum  of  $22  and  $10,  or   $32, 
$32  is  what  per  cent,  of  $4000  ? 

(4)  A  6o-day  draft  for  $5650  costs  $5647.175  ;  ex- 
change being  one  oer  cent,  premium,  what  is  the  rate 
of  interest? 


Methods  in  Written  Arithmetic.  175 

The  premium  is  $56.50.  The  interest,  then,  is 
$56.50  plus  the  difference  between  $5650  and  $5647. 
175,  which  is  ^2.825.  $56-5°+$2-825=#59-325- 

This  sum  is  the  interest  on  $5650  for  63  days.  The 
interest  for  3  days  would  be  1-21  of  this  amount,  and 
for  360  days  would  be  120  times  that  result.  The 
work  may  be  expressed  as  follows : 


21 

Cancelling  and  multiplying,  the  result  is  £339. 
The   question   now  is,  $339  is  what  per  cent,  of 
£5650? 

What  is  it? 

ARBITRATION    OF   EXCHANGE. 

Define  Circular  Exchange.  What  is  its  advan- 
tage? 

Define  the  Arbitration  of  Exchange. 

The  following  problems  from  Ray's  Higher  Arith 
metic  will  illustrate  the  subject : 

i.  A  Louisville  merchant  has  $10,000  due  him  in 
Charleston.  Exchange  on  Charleston  is  ^  per  cent, 
premium.  Instead  of  drawing  directly,  he  advises 
his  debtor  to  remit  to  his  agent  in  New  York  at  $ 
per  cent,  premium,  on  whom  he  immediately  draws 


176  Methods  in   Written  Arithmetic, 

at  12  da.,  and  sells  the  bill  at  i^  per  cent,  premium, 
interest  off  at  6  per  cent.  What  does  he  realize  in 
this  way,  and  what  gain  over  the  direct  exchange  ? 

The  Louisville  merchant  has  $10,000  to  his  credit 
in  Charleston.  He  may  draw  a  draft  on  his  debtor 
and  sell  it  at  %  per  cent,  premium,  receiving  for  it 
$10,025. 

He  advises  his  debtor,  instead,  to  invest  the 
$10,000  in  New  York  exchange  which,  in  Charleston, 
is  at  ^£  per  cent,  premium.  The  $10,000  will  buy  a 
sight  draft  on  New  York  for  $9962.64.  (How  found?) 
This  draft  is  sent  to  the  New  York  agent  of  the  Louis- 
ville merchant.  The  credit  is  now  transferred  from 
Charleston  to  New  York,  where  the  merchant  has  a 
credit  of  $9962.64.  He  now  draws  a  12  day  draft  on 
his  agent  for  that  amount,  and  at  once  sells  it  at  ij^ 
per  cent,  premium  less  the  bank  discount  for  15  days 
at  6  per  cent,  interest. 

The  bank  discount  is  %  per  cent,  of  the  face  of 
the  draft. 

The  net  premium,  then,  is  i%  per  cent.;  he  con- 
sequently receives  for  his  draft  101^  per  cent,  of 
$9962.64.  What  is  the  gain  over  the  first  method? 

2.  A  merchant  of  St.  Louis  owes  $7165.80   to  a 


Methods  in  Written  Arithmetic. 


177 


Baltimore  merchant,  the  amount  being  due  in  St. 
Louis.  Exchange  on  Baltimore  is  ^  per  cent,  pre- 
mium ;  but,  on  New  Orleans  it  is  ^  per  cent,  pre- 
mium ;  from  New  Orleans  to  Havana  ^5  per  cent, 
discount ;  from  Havana  to  Baltimore  ^  per  cent, 
discount.  What  will  be  the  value  in  Baltimore  by 
each  method,  and  how  much  better  is  the  circular  ? 

If  the  St.  Louis  merchant  invests  the  $7165.80  in 
Baltimore  exchange  he  will  purchase  a  draft  whose 
face  is  $7147.93,  (How  obtained?)  and  send  it  to  his 
creditor. 

He  is  directed  to  buy  a  draft  on  New  Orleans,  y& 
per  cent,  premium. 

With  $7165.80  he  can  buy  a  draft  whose  face  is 
$7156.85. 

This  draft  is  sent  to  a  bank  in  New  Orleans  with 
directions  to  issue  a  draft  on  Havana  for  the  amount 
that  it  will  purchase. 

Havana  exchange  being  ^  per  cent,  discount 
$7156.85  will  buy  a  draft  whose  face  is  $7165.81. 
This  draft  is  sent  to  a  Havana  bank  with  direc- 
tions to  invest  its  proceeds  in  Baltimore  exchange, 
which  is  ^  per  cent,  discount.  $7165.81  will  buy 
a  Baltimore  draft  whose  face  is  $7183.77.  This  draft 


178  Methods  in  Written  Arithmetic. 

is  sent  to  the  Baltimore  creditor,  who  presents  it  and 
receives  payment. 

What  is  his  gain  by  the  circular  method? 

The  following  method  is  shorter,  as  it  employs  can- 
cellation. 

Since  exchange  on  New  Orleans  is  *&  per  cent, 
premium,  its  cost  is  100^}  per  cent.,  or  801-100  of 
its  par  value.  The  face  of  the  draft,  then,  is  800- 
801  of  its  cost.  $7165,80  will  buy  a  draft  on  New 
Orleans  whose  face  is  800-801  of  $7165.80. 

Since  exchange  on  Havana  is,  in  New  Orleans,  at 
}i  per  cent,  discount,  its  cost  is  99^6  per  cent.,  oi 
799-800  of  its  face.  Its  face,  consequently,  is  800- 

799  of  its  cost,  hence  the  New  Orleans   draft  will 
purchase  a  draft  on  Havana  for  800-799  °f  801-800 
of  $7165.80. 

Since  Baltimore  exchange  is,  in  Havana,  at  %  per 
cent,  discount,  its  cost  is  99^  per  cent.,  or  399-400 
of  its  face,  hence  its  face  is  400-399  of  its  cost. 

The  Havana  draft,  therefore,  will  purchase  a  Bal- 
timore draft  whose  face  is  400-399  of  800-799  °f  801- 

800  of  $7165.80. 

This  expression  reduced  will  give  the  face  of  the 
Baltimore  draft  that  can  be  obtained  by  the  circular 
method. 


CHAPTER  XXIII. 


EQUATION   OF   PAYMENTS. 

The  Equation  of  Payments  is  the  process  of  find- 
ing a  time  at  which  several  sums  due  at  different 
times,  and  not  bearing  interest,  can  be  paid,  without 
loss  to  debtor  or  creditor,  and  without  the  transfer 
of  money  .for  interest. 

If  the  several  sums  bear  interest  all  could  be  paid 
at  any  time  by  discharging  principal  and  accrued  in- 
terest, and  no  one  would  lose. 

The  principle  upon  which  the  operations  are  bas- 
ed is  very  simple :  If  money  is  paid  before  it  is 
due  interest  should  be  allowed  upon  it :  if  not  paid 
until  after  it  is  due  it  should  bear  interest  from  ma- 
turity until  date  of  payment 

PROBLEM. 

"  I  owe  $500,  due  in  four  months ;  $600,  due  in 
seven  months,  and  $1000  due  in  nine  months  ;  what 

is  the  average  term    of  credit  ?" 
(179^ 


iSo  Methods  in  Written  Arithmetic. 

These  sums  do  not  bear  interest  until  after  matur- 
ity. If  I  should  pay  any  one  of  them  in  full  before 
its  maturity  I  should  lose  the  use  of  it  for  the  given 
time.  If  I  should  not  pay  at  maturity  the  creditor 
would  be  the  loser. 

I  first  experiment  a  little.  Suppose  I  should  pay 
them  in  full  to-day ;  assuming  money  to  be  worth  six 
per  cent.,  what  should  I  lose?  On  the  first  I  should 
lose  the  interest  for  four  months,  or  $10;  on  the  sec- 
ond $21,  and  on  the  third  $45.  This  is  obviously 
true,  for  if  I  had  the  money  in  hand  I  could  lend  the 
different  amounts  for  the  several  periods  and  realize 
the  $76  as  interest  before  the  maturity  of  the  debts. 
If  I  pay  the  debts  to-day  I  shall  pay  out  $2100.  If  I 
retain  this  sum  until  it  earns  me  $76  at  six  per  cent, 
and  then  pay  it  to  my  creditor,  neither  will  lose. 
The  interest  on  $2100  for  one  day  at  6  per  cent,  is  35 
cents.  I  should  retain  it  as  many  days  as  $67  is 
times  3£  cents. 

It  is  not  very  material  what  day  I  select  with 
which  to  make  the  "experiment."  Select  the  most 
convenient  date  visible  by  simple  inspection. 

A  problem  in  Average  of  Accounts  is  simply  a 
double  problem  in  Equation  of  Payments.  First  see 
that  the  pupils  understand  the  meaning  of  the  pro- 


Methods  in  Written  Arithmetic.  181 

blem.  It  may  be  a  transcript  of  a  ledger  account, 
and,  consequently,  not  understood.  The  left-hand 
side  shows  amounts  due  the  owner  of  the  account 
from  one  whose  name  stands  at  the  head  of  it,  while 
the  right-hand  side  shows  the  amounts  due  the  one 
whose  name  is  at  the  head,  from  the  owner  of  the 
account. 

Beginning  with  the  debtor  side,  select  some  date 
and  proceed  as  in  the  problem  given  so  far  as  to  as- 
certain the  loss  or  gain  if  all  be  paid  on  the  date  se- 
lected Taking  the  same  date  perform  the  same  ex- 
periment with  the  credit  side. 

Suppose  the  sum  of  the  amounts  on  the  debtor 
side  to  exceed  the  sum  on  the  credit  side  by  $100. 
This  balance,  then,  is  due  the  owner  of  the  account. 
Suppose  the  interest  items  on  the  credit  side,  how- 
ever, to  exceed  the  similar  items  on  the  debtor  side 
by  $10.  The  meaning  should  be  obvious.  If  the 
balance  of  the  account  be  paid  on  the  "trial  date," 
the  owner  of  the  account  will  lose  $10  of  interest. 
Is  the  trial  date  too  early  or  too  late  ?  How  much, 
if  the  money  is  worth  6  per  cent.? 


CHAPTER  XXIV. 


EVOLUTION. 

The  only  point  under  this  head  that  will  be  dis- 
cussed is  the  formation  of  the  trial  divisor  in  the  ex- 
traction of  cube  root. 

The  rule  as  ordinarily  stated  is  as  follows : 

"Take  three  -times  the  square  of  the  root  already 
found,  as  a  trial  divisor." 

By  this  method  the  work  becomes  very  tedious  if 
there  are  several  terms  in  the  root.  Instead  of  the 
above,  use  the  following : 

Below  the  completed  divisor  write  the  square  of 
the  last  term  of  the  root.  Find  the  sum  of  this 
square,  the  completed  divisor,  and  the  two  additions 
made  to  the  preceding  trial  divisor. 

The  following  will  illustrate  the  plan : 
(182) 


Methods  in   Written  Arithmetic. 


•V    869,457,256 


729 


24300        I40572 
135° 

25 

128375 


25675 


25        12197256 


27075 

By  the  old  method  the  trial  divisor  would  be  found 
by  squaring  95  and  multiplying  it  by  3,  by  the  pres- 
ent, all  that  is  necessary  is  to  write  25  below  25675, 
and  find  the  sum  of  the  25,  the  25675,  the  25  and 
the  1350. 

Pupils  familiar  with  algebraic  formular  will  readily 
understand  the  rule  by  examining  the  following : 

(a+b+c)8=a3+(3a2+3ab+b'0b+[3(a+b)2+3(a+b) 


The  first  completed  divisor  is  3a2-f  3ab+b2. 
next  trial  divisor  is  3(a+b)2,  or~3a2-f6ab+3b*. 
quantity  exceeds  the  former  by  3ab+2bJ. 


T°4  Methods  in   Written  Arithmetic. 

In  the  given  problem  25675  =3a°-f3'*1H-k2-  b2=25_ 
3ab  =  1350- 

If,  then,  t^  zefasi  we  a<*A  the  25  below  it,  the  25 
above  it,  and  the  1350,  we  have  increased  it  by  zbM- 


This  book  is  DUE  on  the  last  date  stamped  below 


W>R  4      1932 

SEP  7    'IWl 
AUG  4     1953 

%£2  1951 
23  115$ 

MAR  4     1960 


Form  L-9-35»i-8,"28 


135     L:etho 
C77rn  writ 
1883  metic. 

AOG  4     1933 


of  CALIFORNIA 


